v = Vi - g t
when v = 0, the ball is at the top.
0 = 16 - 9.8 t
t = 1.633 seconds to the top
y = Yi + Vi t - (1/2)g t^2
y = 20 + 16*1.633 -4.9 * (1.633)^2
y = 20 + 26.13 - 13.07
y = 33.1 m
A ball is thrown straight upward (along the y-axis) with an initial velocity of 16 m/s from the top of a building located 20 m above ground.
Find the maximum height of the ball. In other words, find the ball's maximum distance from the ground.
Be aware that once you find the maximum displacement by using the equations of motion, you will then add 20 m to your answer. This is because the ball was 20 m high at the start.
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