A rock is thrown upward with a velocity of 15 meters per second from the top of a 49 meter high cliff, and it misses the cliff on the way back down. When will the rock be 3 meters from the water, below? Round your answer to two decimal places.

Answer 1

a = first derivative of the velocity = dv/dt = -9.8 m/s² (negative since gravity is slowing the rock)

Therefore,

-9.8 m/s² = dv/dt, which implies that v(t) = -9.8t m/s + c

Since the rock is thrown up with an initial velocity of 12 m/s, v(0) = 12 m/s

Therefore,

12 m/s = v(0) = -9.8(0) m/s + c, which implies that c = 12 m/s

Therefore,

v(t) = -9.8t m/s + 12 m/s

The velocity is the first derivative of the motion, s(t), so

ds/dt = v(t) = -9.8t m/s + 12 m/s

Therefore,

s(t) = -4.9 t² m + 12t m + C

Since the rock is initally thrown up from a 24 meter high cliff, s(0) = 24m, so

24m = s(0) = -4.9(0)² m + 12(0) m + C

C = 24m

Therefore,

s(t) = -4.9 t² m + 12t m + 24 m

To solve the problem, simply set s(t)=7 and solve for t by the quadratic formula. You will probably get a negative result and a positive result. Be sure to reject the negative result since the ball was not released in negative time.

Answer 2

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