A rock is thrown off a 25 m cliff at 60m/s. It was thrown 32° above the ground How long will the rock be in the air before it hits the ground?

The rock has a height y at time t according to

y = 25 m + (60 m/s) sin(32°) t - 1/2 gt ²

where g = 9.8 m/s² is the magnitude of the acceleration due to gravity.

Solve for t when y = 0. The quickest way to do this is with the quadratic formula.

0 = 25 m + (60 m/s) sin(32°) t - 1/2 gt ²

t = (-(60 m/s) sin(32°) + √[(60 m/s)² sin²(32°) - 4 (25 m) (-1/2 gt ²)]) / 2

(ignoring the negative root because that would make t negative)

or approximately

t ≈ 7.2 s

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distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000