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a rock is thrown upward with a velocity of 21 meters per second from the top of a 44 meter high cliff, and it misses the cliff on the way back down. when will the rock be 5 meters from the water, below? round your answer to two decimal places.

The answer to your question is: t = 3.54 s

Step-by-step explanation:

data

vo = 21 m/s

height of the cliff = 44 m

t= ? when the rock be 5 meters from the water.

Process

First, calculate the distance the rock reach in its way up.

    h max = vo² /2g = (21)² / 2(9.81) = 22.5 m

total height = 44 m + 22.5 m = 66.5 m

but we need to calculate the time when the rock be 5 m from the water so,

    height = 66.5 - 5 m

   height = 61.5 m

Now, use free fall formulas to calculate time

   h = gt² / 2

   61.5 = (9.81)t² / 2

solve for t

  2(61.5) = 9.81t²

   123   = 9,81t²

   t² = 123 / 9.81

   t² = 12.53

     t = 3.54 s

(-1,-0.5)

step-by-step explanation:

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