A heavy rock is shot upward from the edge of a vertical cliff. It leaves the edge of the cliff with an initial velocity of 12 m/s directed at 26° from the vertical and experiences no appreciable air resistance as it travels. The distance the rock landed was 40.4 m away from the cliff. How high is the cliff?

The height of the building is 88.63 m.

Explanation:

Given;

initial component of vertical velocity, = 12 m/s sin 26° = 5.26 m/s

initial horizontal component of the velocity, = 12 m/s cos 26° =10.786 m/s

horizontal distance traveled by the rock, x = 40.4 m

time of flight is calculated as;

x = t

t = x /

t = 40.4 / 10.786

t = 3.75 s

Determine the final vertical velocity of the ball;

Determine the height of the rock;

Therefore, the height of the building is 88.63 m.

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