Question

An archer shoots an arrow at a 75.0 m distant target; the bull's-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull's-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

Answer 1

(a) $18.4^{\circ}$

We know that the horizontal distance travelled by the arrow is

d = 75.0 m

We also know that the horizontal range of a projectile is given by

$d=\frac{v^2}{g} sin 2\theta$

where

v is the speed of the projectile

g = 9.8 m/s^2 is the acceleration of gravity

$\theta$ is the angle of the projectile

Here we have

v = 35.0 m/s

Substituting into the equation and solving for $\theta$, we find

$\theta=\frac{1}{2}sin^{-1} (\frac{dg}{v^2})=\frac{1}{2}sin^{-1} (\frac{(75.0 m)(9.8 m/s^2)}{(35.0 m/s)^2})=18.4^{\circ}$

(b) 6.2 m

In order to answer this part of the problem, we have to calculate what is the maximum height reached by the projectile in its trajectory.

First of all, we can calculate the vertical component of the velocity, which is given by:

$u_y = u sin \theta = (35.0 m/s) sin 18.4^{\circ} = 11.0 m/s$

The motion along the vertical direction is a uniformly accelerated motion with constant acceleration

g = -9.8 m/s^2

(negative since it points downward). So we can write

$v_y^2 - u_y^2 = 2gh$

where

$v_y = 0$ is the vertical velocity at the point of maximum height

h is the maximum height

Solving for h, we find

$h=\frac{v_y^2 - u_y^2}{2g}=\frac{0-(11.0 m/s)^2}{2(-9.8 m/s^2)}=6.2 m$

Therefore, the arrow will go over the branch (which is located 3.50 m above the ground).