A ball is thrown straight upward and returns to the thrower's hand after 1.8 s in the air. A second ball is thrown at an angle of 23◦ with the horizontal. At what speed must the second ball be thrown so that it reaches the same maximum height as the first? The acceleration of gravity is 9.8 m/s 2. Answer in units of m/s.

U = 9.1 m/s

Explanation:

from the question we are given the following

time (t) = 1.8 s

angle = 23 degrees

acceleration due to gravity (g) = 9.8 m/s^{2}

let us first calculate the initial velocity (u) which too the first ball to its maximum height from the equation below

v = u + 0.5at

The final velocity (v) is zero since the ball comes to rest The time (t) it takes to get to the maximum height would be half the time it is in the air, t = 0.5 x 1.8 = 0.9

therefore

0 = u - (0.5 x 9.8 x 0.9)

u = 7.9 m/s

for the second ball to get to the maximum height of the first ball, the vertical component of its initial velocity (U) must be the same as the initial velocity of the first ball. therefore

U sin 60 = 7.9

U = 7.9 : sin 60

U = 9.1 m/s