Horizontal Range (arrow fired max speed off moving horse)

[Cascade]

Homework Statement

A standing archer can fire an arrow from a bow with a maximum speed of 40.0 m/s. The archer rides a horse that can run at 12.0 m/s. The archer shoots an arrow at the maximum speed while riding the horse in the direction the horse is running and 10 degrees above horizontal. The arrow is released 2.35 m above the ground. What is the horizontal range of the arrow from the point the archer releases the arrow?

Homework Equations

[1] $$y - y_0 = v_0 t + \frac{1}{2} a t^2$$
[2] $$\frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a}$$
[3] $$x = v_x t$$

The Attempt at a Solution

Let the initial vertical height = 2.35m & the final vertical height = 0m & a = -g = -9.81m/s^2
Find the initial vertical velocity.
$$v_0 = (40m/s)sin(10) = 6.946m/s$$
Plug into equation [1], multiply by -1, re-arrange, then use quadratic formula [2] to find the time.
$$0m - 2.35m = (6.946m/s) t - (4.905m/s^2) t^2 => 0m = (4.905m/s^2) t^2 -6.946m/s - 2.35m$$
$$t = \frac{-(-6.946) \pm \sqrt{(-6.946)^2 - 4 (4.905) (-2.35)}}{2 (4.905)} = -0.282s~ or~ 1.698s$$

Now this is where I get confused. Do I take into account the horse's velocity when finding the horizontal velocity?
Find the horizontal velocity.
$$v_x = (40m/s)cos(10) + 12.0m/s = 51.392m/s$$
Find the horizontal range with equation [3].
$$x = (51.392m/s) (1.698s) = 87.264m$$

 

[haruspex]

That all looks right, except you quote too many significant digits in the answer.

 

[SteamKing]

Homework Statement

A standing archer can fire an arrow from a bow with a maximum speed of 40.0 m/s. The archer rides a horse that can run at 12.0 m/s. The archer shoots an arrow at the maximum speed while riding the horse in the direction the horse is running and 10 degrees above horizontal. The arrow is released 2.35 m above the ground. What is the horizontal range of the arrow from the point the archer releases the arrow?

Homework Equations

[1] $$y - y_0 = v_0 t + \frac{1}{2} a t^2$$
[2] $$\frac{-b \pm \sqrt{b^2 - 4 a c}}{2 a}$$
[3] $$x = v_x t$$

The Attempt at a Solution

Let the initial vertical height = 2.35m & the final vertical height = 0m & a = -g = -9.81m/s^2
Find the initial vertical velocity.
$$v_0 = (40m/s)sin(10) = 6.946m/s$$
Plug into equation [1], multiply by -1, re-arrange, then use quadratic formula [2] to find the time.
$$0m - 2.35m = (6.946m/s) t - (4.905m/s^2) t^2 => 0m = (4.905m/s^2) t^2 -6.946m/s - 2.35m$$
$$t = \frac{-(-6.946) \pm \sqrt{(-6.946)^2 - 4 (4.905) (-2.35)}}{2 (4.905)} = -0.282s~ or~ 1.698s$$

Now this is where I get confused. Do I take into account the horse's velocity when finding the horizontal velocity?
Find the horizontal velocity.
$$v_x = (40m/s)cos(10) + 12.0m/s = 51.392m/s$$
Find the horizontal range with equation [3].
$$x = (51.392m/s) (1.698s) = 87.264m$$

When it is fired, the arrow leaves the bow with a velocity relative to the bow, not to the ground.

When he fires the arrow, the archer is moving with the horse.

You can't arbitrarily say that the vertical velocity of the arrow as it leaves the bow is the same as if the archer is standing still on the ground, while at the same time including the horse's speed just for the horizontal motion.

 

[haruspex]

When it is fired, the arrow leaves the bow with a velocity relative to the bow, not to the ground.

When he fires the arrow, the archer is moving with the horse.

You can't arbitrarily say that the vertical velocity of the arrow as it leaves the bow is the same as if the archer is standing still on the ground, while at the same time including the horse's speed just for the horizontal motion.

It depends what you think the 10 degree launch angle means. I took it as the angle as aimed by the archer. Are you suggesting it should be the initial trajectory as viewed by a bystander?

 

[SteamKing]

It depends what you think the 10 degree launch angle means. I took it as the angle as aimed by the archer. Are you suggesting it should be the initial trajectory as viewed by a bystander?

The OP clearly states that the arrow is fired 10 degrees above the horizontal. One of the reasons to do that is to increase the range of the arrow, just like shooting the arrow from a moving horse increases the range.

 

[haruspex]

The OP clearly states that the arrow is fired 10 degrees above the horizontal.

Yes, but what does that mean? Is the launch angle 10 degrees above horizontal as viewed by the archer, or 10 degrees above horizontal as viewed by a bystander on the ground? They are not the same.
It seems to me it surely means the angle at which the archer aimed, so is the launch angle from the archer's perspective. That being so, the OP calculation is correct. Think of it in the reference frame of the archer.

 

[SteamKing]

Yes, but what does that mean? Is the launch angle 10 degrees above horizontal as viewed by the archer, or 10 degrees above horizontal as viewed by a bystander on the ground? They are not the same.
It seems to me it surely means the angle at which the archer aimed, so is the launch angle from the archer's perspective. That being so, the OP calculation is correct. Think of it in the reference frame of the archer.

I don't see how the archer can determine that he is firing his arrow 10 degrees above the horizontal without making reference to the horizontal at some point. IMO, the only way the vertical velocity is not affected by the motion on horseback is if the horse itself is standing still when the archer fires arrow.

I'm afraid we're going to have to agree to disagree on this point.

 

[haruspex]

I don't see how the archer can determine that he is firing his arrow 10 degrees above the horizontal

To launch it at 10 degrees above the horizontal from the archer's perspective is easy. The archer just tilts the bow up 10 degrees. What the archer would find hard is launching it at 10 degrees relative to the ground. To do that, the archer would have to aim higher.
As I posted, just consider the problem in the reference frame of the archer. Calculate how far in front of the archer the arrow will land, then add the distance the archer has ridden.

 

[polo101]

This may sound silly , but does it mean that the energy of the arrow fired by a horse archer increased comparing to a standing still archer ? As we all know energy cannot be created or destroyed, and the kinetic energy equation is 1/2 mv^2 . If the velocity of the arrow increased, does it mean that the KE of the arrow increased ?

 

[jbriggs444]

This may sound silly , but does it mean that the energy of the arrow fired by a horse archer increased comparing to a standing still archer ? As we all know energy cannot be created or destroyed, and the kinetic energy equation is 1/2 mv^2 . If the velocity of the arrow increased, does it mean that the KE of the arrow increased ?

Energy is conserved. That means that once you choose a reference frame, the total energy, as measured according to that frame, will remain the same. Energy is not invariant. That means that if you change reference frames, you change the assessment of the starting total energy. And you may change your assessment of where the energy is within the system.

Put that to one side. We have an archer who shoots an arrow using a certain amount of chemical potential energy in his muscles to do so. As measured from one frame of reference this increases the energy of the arrow by one amount. As measured from another frame of reference, it increases the energy of the arrow by a different amount. How can this be?

The answer is in Newton's third law. Firing the arrow involved two forces. The force of the archer on the arrow and the force of the arrow on the archer. If we use a frame of reference in which the archer is stationary then the archer does not move [much] under the force of the arrow. Accordingly, that force does no work. But if we adopt a frame in which the archer is moving then the arrow is exerting a force opposite to the direction of motion. It does negative work. In the end it all balances out and the net increase in kinetic energy matches the decrease in chemical potential energy in the archer's muscles no matter which frame of reference we use. Conservation of energy is upheld. Nonetheless, more energy is injected into the arrow if the archer is considered to be moving.