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?

Solution 29PE

Step 1 of 6</p>

An arrow is released at an angle with a speed towards a bull’s eye which is at the same height as the release height of arrow and at a horizontal distance R=75 m from the arrow release position. Here we need to calculate the angle at which the arrow has to be released, so that it hits the bull’s eye. In the second part we have to find, at the half way between the release point and the target is the arrow is moving above or below a branch of height 3.5m.

Step 2 of 6</p>To determine the angle at which arrow has to be released,

We can calculate the angle with which arrow has to be released by using range equation of projectile motion and it is given by

Range equation is,

Solving for angle ,

Step 3 of 6</p>

Substituting , R=75 m, in above equation

Therefore, in order to hit the bull’s eye, the arrow has to be released at above the horizontal axis at speed 35 m/s.