At what speed must the second ball be thrown so that it reaches the same maximum height as the first?
A ball is thrown straight upward and returns to the thrower's hand after 4.1s in the air. A second ball is thrown at an angle of 67 degrees with the horizontal.
The acceleration of gravity is 9.8m/s^2. Answer in units of m/s
The first ball is thrown straight up and returns to the thrower's hand in 4.1 s. The time take by the ball to reach the highest point is half the total time traveled by the ball. That is equal to 2.05 s. If the ball was thrown up with an initial velocity of V, the velocity at the highest point after 2.05 s is equal to 0.
=> 0 = V - 9.8*2.05
=> V = 20.09 m/s
The height of the ball thrown at 20.09 m/s after 2.05 s is 20.09*2.05 - (1/2)*9.8*(2.05)^2 = 20.59 m
The second ball is thrown at an angle of 67 degrees with the horizontal. If the velocity of the ball is V, the vertical component of the velocity is V*sin 67.
At the highest point the vertical velocity is equal to 0
=> 0^2 - (V*sin 67)^2 = -2*9.8*20.59
=> V^2 = 2*9.8*20.59/(sin 67)^2
=> V = sqrt ( 476.27)
=> V = 21.82 m/s
The ball must be thrown at 21.82 m/s so that it reaches the same maximum height as the first.