Answer:
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°.
