What Angle Should the Archer Shoot the Arrow for Optimal Accuracy?

[xxpbdudexx]

Usually these problems are really simple for me, but this one is making me lose my mind:

"An archer wishes to shoot an arrow at a target at eye level a distance of 49.0 m away. If the initial speed imparted to the arrow is 69.3 m/s, what angle should the arrow make with the horizontal as it is being shot?"

theta = cos^-1(v_0x/v₀)

v_0x=d/t

I just have no idea how to find t. Help please?

 

[Curious3141]

Usually these problems are really simple for me, but this one is making me lose my mind:

"An archer wishes to shoot an arrow at a target at eye level a distance of 49.0 m away. If the initial speed imparted to the arrow is 69.3 m/s, what angle should the arrow make with the horizontal as it is being shot?"

theta = cos^-1(v_0x/v₀)

v_0x=d/t

I just have no idea how to find t. Help please?

Consider the vertical motion and use [itex]v = u + at[/itex] where u is initial velocity, v is final velocity, a is acceleration and t is the time. Remember that when the arrow reaches the target, it's at eye level again (level horizontally with the point from which it was shot). Therefore, by the symmetry of the parabolic motion, its final vertical velocity is equal in magnitude but opposite in direction to the initial vertical velocity. Using that, solve for t, substitute into what you already have and use a trig identity to simplify before solving for θ.

 

[xxpbdudexx]

Consider the vertical motion and use [itex]v = u + at[/itex] where u is initial velocity, v is final velocity, a is acceleration and t is the time. Remember that when the arrow reaches the target, it's at eye level again (level horizontally with the point from which it was shot). Therefore, by the symmetry of the parabolic motion, its final vertical velocity is equal in magnitude but opposite in direction to the initial vertical velocity. Using that, solve for t, substitute into what you already have and use a trig identity to simplify before solving for θ.

[itex]v = -u[/itex] where [itex]u = 69.3 m/s[/itex]. So:

[itex]69.3 = -69.3 +at[/itex]

I don't know the resultant acceleration so I assume I am doing something wrong in coming up with this equation from what you have said.

 

[Curious3141]

[itex]v = -u[/itex] where [itex]u = 69.3 m/s[/itex]. So:

[itex]69.3 = -69.3 +at[/itex]

I don't know the resultant acceleration so I assume I am doing something wrong in coming up with this equation from what you have said.

No, 69.3m/s is the actual initial speed in the direction of launch.

What you need is the vertical component. It involves trigonometry, just like how you found the horizontal component.

When you throw an object vertically upward at a velocity u, it will rise to a maximum height then descend. When it passes your hand, it will be traveling downward with velocity -u. You can see this is so from [itex]v^2 = u^2 + 2as[/itex]. When the object is at the level of your hand, the displacement s = 0, so [itex]v = \pm u[/itex], where [itex]v = u[/itex] occurs at the start, and [itex]v = -u[/itex] when the object passes by your hand on its downward trajectory.

This amounts to the same thing when you consider the vertical motion alone. Except that you need to properly resolve the initial velocity into vertical and horizontal components.

Start by drawing a right angled triangle, with the hypotenuse being represented by the initial velocity at an angle [itex]\theta[/itex] to the horizontal. What are the lengths of the other two sides? They represent the respective vertical and horizontal velocities.

 

[asteriatic]

Hello there,

I understand that you are having difficulty finding the time (t) in this problem. Don't worry, let's break it down step by step.

First, we need to understand that the arrow's motion can be divided into two components: horizontal and vertical. The horizontal component is given by v0x, which is the initial speed imparted to the arrow. The vertical component is given by v0y, which is the speed of the arrow in the vertical direction.

Next, we can use the equation v0x = d/t to find the time (t). We know that the distance (d) is 49.0 m and the initial speed (v0x) is 69.3 m/s. Therefore, we can rearrange the equation to t = d/v0x. Substituting the values, we get t = 49.0 m / 69.3 m/s = 0.707 seconds.

Now that we have the time (t), we can use the equation v0y = gt to find the vertical component of the initial speed (v0y). Here, g is the acceleration due to gravity, which is equal to 9.8 m/s^2. Substituting the values, we get v0y = 9.8 m/s^2 * 0.707 s = 6.92 m/s.

Finally, we can use the equation theta = cos^-1 (v0x/v0) to find the angle (theta) at which the arrow should be shot. Here, v0 is the initial speed of the arrow, which is 69.3 m/s. Substituting the values, we get theta = cos^-1 (69.3 m/s / 69.3 m/s) = cos^-1 (1) = 0 degrees.

Therefore, the angle at which the arrow should be shot is 0 degrees with the horizontal. I hope this helps clear up your confusion. Keep practicing and you will become a pro at 2-D kinematics problems in no time.

Best of luck!