A ball is thrown directly upwards from a 40.0 m tall building with a speed of 36

Question

A ball is thrown directly upwards from a 40.0 m tall building with a speed of 36.0 m/s. It reaches max height then falls down, missing the edge of the building and continues to fall, hitting the ground. a. Find the maximum height reached by the ball and the time to reach max height b. Find the velocity of the ball when it returns to 40.0 m, and the time when this occurred c. Find the total time in the air and the impact velocity of the ball d. Find the position of the ball relative to the ground at t = 2.50 s after the toss e. What is the velocity of the ball and its direction of the motion at t = 2.50 s

Explanation / Answer

a] maximum height = 40 + v^2/2g = 40+36^2/19.6 = 106.1 m

time t = v/g = 36/9.8 = 3.673 s

b] again when it returns to same level , v = 36 m/s

time t = 2v/g = 2*36/9.8 = 7.346 s

c] total time T = 3.673 + sqrt(2*106.1/9.8)

= 8.326 s

impact velocity = g (T-3.673) = 9.8*(8.326-3.673)

= 45.6 m/s answer

d] h = h0 + ut - 0.5gt^2 = 40 + 36*2.5 - 0.5*9.8*2.5^2

= 99.4 m

e] v = u - gt = 36 - 9.8*2.5 = 11.5 m/s upward

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