A golfer hits a ball on a level fairway at an angle of 40o relative to the horizontal, and it lands on the green. Another golfer hits a ball with the same speed but with another angle, and it lands on the green by the other ball. Assume no air resistance, what is the launch angle for the second golfer.

θ' = 50°

Explanation:

Given:

- First ball launched with angle θ = 40 °

- The both balls land at the same spot on green

- Both balls hit with same velocity

Find:

what is the launch angle for the second golfer θ'.

Solution:

- We are given that both balls land at the same horizontal distance from initial position. So we can say that the range of projectile motion for both balls is same.

- The range of projectile motion as a function of θ is given as:

                           Range = v^2*sin (2*θ) / g

Where,  v: The initial striking velocity

             g: The gravitational acceleration.

- Since, the Ranges for both balls is same we have:

                           v^2*sin (2*θ) / g = v^2*sin (2*θ') / g

- Velocity v and g are constants and canceled out:

                              sin (2*θ) = sin (2*θ')

- Plug θ = 40° :

                              sin ( 80 ) = sin (2*θ')

                              arcsin ( sin ( 80 ) ) = 2*θ'

                              2*θ' = (80 , 180 - 80)

                              θ' = (40° , 100/2 = 50°)

- Hence, the second ball is launched with a different launch angle θ' = 50°.

answer( sicar );