An archer shoots an arrow at a 22^{o} angle to the horizontal. At the top of its motion, the arrow just clears a 36 m high oak tree. How fast was the arrow going when it left the bow? (Assume it was shot from a height of 1.0 m above the ground.) How far away is the archer from the tree?
In this problem, you are asked to describe the motion (how fast does it fly, how far does it go) of an arrow. Whenever you are asked to describe the motion of an object without worrying about the cause of that motion, you have a kinematics problem.
There are three key kinematic equations. If you carefully select the equation which most directly describes the situation in your problem, you will not only solve the problem in fewer steps but also understand it better. The three equations, written for motion in the y-direction, are:
1. y = y_{0} + v_{0y}Δt + ½ a_{y}(Δt)^{2} (relates position and time)
2. v_{y} = v_{0y} + a_{y}Δt (relates velocity and time)
3. v_{y}^{2} = v_{0y}^{2} + 2a_{y}(Δy) (relates velocity and position)
a) In this problem, you are asked for the initial velocity of the arrow. Initial velocity shows up in all three equations, and so it will not help you select the most useful equation to use. However, you know you want to find the initial velocity that results in the arrow reaching the top of its motion (y-velocity = 0) when it is at a height (y-position) of 36 m. So you will use equation 3 in the y-direction.
b)You are also asked how far the tree is from the archer (x-position). The x-position that you want is the one that corresponds to a y-velocity of zero. Remember, you cannot mix x- and y-components in a single equation.
Whenever you are asked about the x-direction and told about the y-direction, or vice versa, you need to use time to go between the two.
The amount of time that the arrow spends going from the bow to its maximum height (v_{y} = 0) is the same as the amount of time it takes to go from the bow to the tree. So you can use what you know about the y-velocity of the ball to find the time it spends in the air (equation 2 in the y-direction) and then use that time to find its x-position (equation 1 in the x-direction.)
Step 1:
You have answered the first question: the arrow leaves the bow at a speed of 70. m/s. Scroll down to solve for the distance between the archer and the tree.
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Step 2:
Remember that Δt refers to the time between the two points put into the equation. So it takes the arrow 2.7 s to go from the bow to the tree. Now that you know the time it took to cover this distance, scroll down to calculate the distance.
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Step 3:
The problem asked you to find the horizontal (x) distance between the archer (Point 1) and the tree (Point 2.) No further mathematical solution is necessary.
The first question asks you to relate speed and height in the absence of air resistance, so you can, indeed, use energy to solve the first part of the problem. If you do so, remember that kinetic energy at Point 2 is not zero—there is still a horizontal component to the arrow’s motion. You cannot use energy to determine how far away the tree is. That is a horizontal distance, and there is no energy transformation taking place in this problem relating to a horizontal force.
Unless a problem states otherwise, you can assume that air resistance on a small object designed as a projectile is negligible. Therefore, the only significant cause of the arrow’s acceleration is gravity. For all objects near the Earth’s surface, the downward acceleration due to gravity, g, has a value of 9.8 m/s^{2}. (If you have already studied forces, you can see this quickly with a free body diagram.) So the x-component of the arrow’s acceleration is 0, and the y-component is -9.8 m/s^{2}. These values are true throughout the problem and not just at any one point.
The first point described in the problem is when the arrow leaves the bow. I chose that point to be x = 0 and y = 1.0 m. I can pick the origin of the coordinate system anywhere that is convenient, and all other locations are then referenced from that point. With this choice, other horizontal distances in the problem will be relative to the archer, and other vertical distances relative to the ground.
Because the initial velocity is at an angle to the axes, you need to divide it into its x- and y-components.
As you can see in the figure, the x- and y-components of a vector make up the sides of a right triangle. The vector itself forms the hypotenuse (h). The side of the triangle opposite the angle that you use is given by h sinθ and the side that touches the angle you use is given by h cosθ (soh cah toa)
In this case, the x-component is adjacent to the 22^{o} angle, and so can be given by (v_{1})cos22^{o} as shown. Likewise, the y-component is opposite to the 22^{o} angle and is therefore given by (v_{1})sin22^{o}.
22^{0} to the horizontal means that if you start with a horizontal line (in the +x direction) and rotate it counter-clockwise by 22^{0} you will have the correct direction for the initial velocity of the arrow.
The first point described in the problem is when the arrow leaves the bow. I chose that point to be x = 0 and y = 1.0 m. I can pick the origin of the coordinate system anywhere that is convenient, and all other locations are then referenced from that point. With this choice, other horizontal distances in the problem will be relative to the archer, and other vertical distances relative to the ground.
Because the initial velocity is at an angle to the axes, you need to divide it into its x- and y-components.
We know that at the top of an arc the y-component of an object’s motion is zero. (The object slows down due to gravity as it goes up, reaches a stop vertically, and then speeds up in the y-direction as it moves down.) In the absence of air resistance, the x-component of the object’s velocity remains unchanged so the arrow continues to move forward at a constant speed.
In this case, the top of the arc was at a location just above the oak tree, or 36 m above the ground. We are asked to find the x distance from the archer to the tree as part of the problem.
The kinematic variables (s, v, a) are vectors, and therefore you will work with kinematic equations in one dimension (x- or y-) at a time. Therefore, you will work only with x- or y-components of these variables, and so you need to divide them into their components.
The t in the kinematic equations refers to the time interval between the two points in the equation, with y_{0} occurring at the earlier time. I use Δt rather than t to be explicit that this is a time interval (t – t_{0}) and not a point in time.
Some text books will give more than three kinematic equations—for example, they may provide “range equations” or different versions of the equations for horizontal and vertical motion. Those are just these three equations solved for special cases. In my view, it is better to know and understand three equations that will always work than to memorize many equations which are specific to certain situations only.
In this problem, you know two things about Point 2 in the y-direction. You know both position (36 m) and velocity (0), so either the position-time or the velocity-time equation can be used to solve for time. I used the velocity time equation merely because it is simpler. Either equation will give the same answer.
You need to find the starting velocity of the arrow that allows it to go from the archer (Point 1) to the top of the tree (Point 2) at its maximum height. Therefore, “y_{0}” and “y” are “y_{1}” and “y_{2}” respectively.
The acceleration of the ball (due to gravity) is downward, which is the negative y-direction as chosen in this problem. Because a, v, and y are all vector quantities, their values are given a + or – sign to represent direction whenever they are filled into any of the kinematic equations.
Remember to use only y-components of the arrow’s velocity. At the starting point (Point 1), the arrow flies with an unknown velocity at an angle of 22^{0} relative to the horizontal. It’s y-motion becomes zero at the top of the arc, or Point 2.
Regardless of where you pick y = 0, Point 2 is 35 m higher than Point 1.
Only two significant figures were given in the text of the problem, so only two significant figures are included in the solution.
The acceleration of the ball (due to gravity) is downward, which is the negative y-direction as chosen in this problem. Because a, v, and y are all vector quantities, their values are given a + or – sign to represent direction whenever they are filled into any of the kinematic equations.
Remember you are working in the y-direction, so you should only include the y-component of the arrow’s velocity. At Point 1 (the archer), the arrow travels with a speed of 70. m/s at an angle of 22^{o} to the horizontal. At Point 2, the arrow is at the top of its motion and so it’s y-velocity is zero.
You are asked to find how far the arrow travels from the archer (Point 1) to the tree (Point 2). Therefore, “x_{0}” and “x” are “x_{1}” and “x_{2}” respectively.
The acceleration of the baseball in the y-direction is - 9.8 m/s^{2}. However, there are no forces acting on the baseball in the x-direction, and so a_{x} = 0.
Only two significant figures were given in the text of the problem, so only two significant figures are included in the solution.
Remember that you are working in the x-direction, and so you only want to include the x-component of the arrow’s motion.
The first question asks you to relate speed and height in the absence of air resistance, so you can, indeed, use energy to solve the first part of the problem. If you do so, remember that kinetic energy at Point 2 is not zero—there is still a horizontal component to the arrow’s motion. You cannot use energy to determine how far away the tree is. That is a horizontal distance, and there is no energy transformation taking place in this problem relating to a horizontal force.
All that matters for describing motion is that you explicitly state where your reference point is. In this case, you know the initial position of the arrow and the height of the tree relative to the ground, so I chose the ground to be y = 0. Therefore, any heights that I find in the problem will be heights above the ground. I chose x = 0 to be the starting point of the problem—the location of the archer—so any distances that I find will be relative to the archer.
s is the vector symbol for displacement—in other words, the vector whose components are x- and y-. Some books use x and others use r instead of s, so use whichever variable is comfortable for you.
Using a table like this in two dimensional kinematic problems can reduce one of the most common errors that students make on exams. The kinematic variables are vectors, and so you can only work in one dimension at a time. So you need to divide vectors into components. However, you also have three equations in each direction and so you need to think about which equation to use. It is very common for students to go to the work of finding the components for the vectors, but then to use the full vector (in this case, v_{1}) rather than the appropriate component when it is time to solve the equation, or to use -9.8 m/s^{2} for acceleration in the x-direction as well as in y-direction . By organizing your information into a table and working with only the x- or y- column at a time, you are less likely to make these mistakes when you solve the problem.
Unless a problem states otherwise, you can assume that air resistance on a small object designed as a projectile is negligible. Therefore, the only significant cause of the arrow’s acceleration is gravity. For all objects near the Earth’s surface, the downward acceleration due to gravity, g, has a value of 9.8 m/s^{2}. (If you have already studied forces, you can see this quickly with a free body diagram.) So the x-component of the arrow’s acceleration is 0, and the y-component is -9.8 m/s^{2}. These values are true throughout the problem and not just at any one point.
Because the initial velocity is at an angle to the axes, you need to divide it into its x- and y-components.
As you can see in the figure, the x- and y-components of a vector make up the sides of a right triangle. The vector itself forms the hypotenuse (h). The side of the triangle opposite the angle that you use is given by h sinθ and the side that touches the angle you use is given by h cosθ (soh cah toa)
In this case, the x-component is adjacent to the 22^{o} angle, and so can be given by (v_{1})cos22^{o} as shown. Likewise, the y-component is opposite to the 22^{o} angle and is therefore given by (v_{1})sin22^{o}.
The kinematic variables (s, v, a) are vectors, and therefore you will work with kinematic equations in one dimension (x- or y-) at a time. Therefore, you will work only with x- or y-components of these variables, and so you need to divide them into their components.
The t in the kinematic equations refers to the time interval between the two points in the equation, with y_{0} occurring at the earlier time. I use Δt rather than t to be explicit that this is a time interval (t – t_{0}) and not a point in time.
Some text books will give more than three kinematic equations—for example, they may provide “range equations” or different versions of the equations for horizontal and vertical motion. Those are just these three equations solved for special cases. In my view, it is better to know and understand three equations that will always work than to memorize many equations which are specific to certain situations only.
In this problem, you know two things about Point 2 in the y-direction. You know both position (36 m) and velocity (0), so either the position-time or the velocity-time equation can be used to solve for time. I used the velocity time equation merely because it is simpler. Either equation will give the same answer.
You need to find the starting velocity of the arrow that allows it to go from the archer (Point 1) to the top of the tree (Point 2) at its maximum height. Therefore, “y_{0}” and “y” are “y_{1}” and “y_{2}” respectively.
The acceleration of the ball (due to gravity) is downward, which is the negative y-direction as chosen in this problem. Because a, v, and y are all vector quantities, their values are given a + or – sign to represent direction whenever they are filled into any of the kinematic equations.
Remember to use only y-components of the arrow’s velocity. At the starting point (Point 1), the arrow flies with an unknown velocity at an angle of 22^{o} relative to the horizontal. It’s y-motion becomes zero at the top of the arc, or Point 2.
Only two significant figures were given in the text of the problem, so only two significant figures are included in the solution.
The acceleration of the ball (due to gravity) is downward, which is the negative y-direction as chosen in this problem. Because a, v, and y are all vector quantities, their values are given a + or – sign to represent direction whenever they are filled into any of the kinematic equations.
Remember you are working in the y-direction, so you should only include the y-component of the arrow’s velocity. At Point 1 (the archer), the arrow travels with a speed of 70. m/s at an angle of 22^{o} to the horizontal. At Point 2, the arrow is at the top of its motion and so it’s y-velocity is zero.
You are asked to find how far the arrow travels from the archer (Point 1) to the tree (Point 2). Therefore, “x_{0}” and “x” are “x_{1}” and “x_{2}” respectively.
The acceleration of the baseball in the y-direction is - 9.8 m/s^{2}. However, there are no forces acting on the baseball in the x-direction, and so a_{x} = 0.
Only two significant figures were given in the text of the problem, so only two significant figures are included in the solution.
Unless a problem states or implies otherwise, you can assume that air resistance on a small object designed as a projectile is negligible. Therefore, the only significant cause of the arrow’s acceleration is gravity. For all objects near the Earth’s surface, the downward acceleration due to gravity, g, has a value of 9.8 m/s^{2}. (If you have already studied forces, you can see this quickly with a free body diagram.) So the x-component of the arrow’s acceleration is 0, and the y-component is -9.8 m/s^{2}.
All that matters for describing motion is that you explicitly state where your reference point is. In this case, you know the initial position of the arrow and the height of the tree relative to the ground, so I chose the ground to be y = 0. Therefore, any heights that I find in the problem will be heights above the ground. I chose x = 0 to be the starting point of the problem—the location of the archer—so any distances that I find will be relative to the archer.
s is the vector symbol for displacement—in other words, the vector whose components are x- and y-. Some books use x and others use r instead of s, so use whichever variable is comfortable for you.
Using a table like this in two dimensional kinematic problems can reduce one of the most common errors that students make on exams. The kinematic variables are vectors, and so you can only work in one dimension at a time. So you need to divide vectors into components. However, you also have three equations in each direction and so you need to think about which equation to use. It is very common for students to go to the work of finding the components for the vectors, but then to use the full vector (in this case, v_{1}) rather than the appropriate component when it is time to solve the equation, or to use -9.8 m/s^{2} for acceleration in the x-direction as well as in y-direction . By organizing your information into a table and working with only the x- or y- column at a time, you are less likely to make these mistakes when you solve the problem.
The kinematic equations relate position and velocity at any two points in the motion. Because many problems provide or ask about information at more than two points, I identify each point with its own subscript to keep them straight. I can then pick any two points to be s_{0} and s when I use the equations. The t in the kinematic equations refers to the time interval between the two points in the equation, with s_{0} occurring at the earlier time. I use Δt rather than t to be explicit that this is a time interval (t – t_{0}) and not a point in time.
In this problem, you are asked to find how fast an arrow needs to travel in order to reach a maximum (y-velocity = 0) height of 36 m. Therefore, regardless of the unknown quantity, the most direct relationship for the problem is the position-velocity equation in the y-direction.
Next, you are asked how far the arrow travels during that time. Displacement, velocity and acceleration are vector quantities, and so you cannot mix x- and y- components in the same equation. Therefore, you cannot use the y-velocity condition (v_{y2} = 0) to solve for the x-position question (how far away is the tree) in a single step.
You can always relate x- and y-motion through time—the arrow spends as long going up as it spends going from the archer to tree. So we first solved for time required to go up to a maximum point (y-velocity) and then used that time to determine how far the ball traveled (x-position.)