Question

A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 3 . 2 s later with an initial speed of 62 . 72 m / s. They hit the ground at the same time.

(a) How long does it take the first stone to hit the ground? The acceleration of gravity is 9 . 8 m / s 2 . Answer in units of s.

(b) How high is the cliff? Answer in units of m.

Answer #1

(a)

For the first rock,

d1 = (vi)t + 1/2at^2

d1 = 1/2gt^2 (since vi = 0)

For the second,

d1 = (vi)t + 1/2at^2

d1 = (62.72)(t - 3.2) + 1/2g(t - 3.2)^2

So, since they fell the same distance,

1/2gt^2 = (62.72)(t - 3.2) + 1/2g(t - 3.2)^2

1/2gt^2 = 62.72t - (3.2 * 62.72) + 1/2g(t^2 - 6.4t + 10.24)

gt^2 = 125.44 - 2(3.2 * 62.72) + gt^2 - 6.4gt + 10.24g

6.4gt = 125.44 - 2(3.2 * 62.72) + 10.24g

t = **6s** for
the first stone, the one with no initial velocity, to hit the
ground.

(b)

Now, the height of the cliff...

d1 = (vi)t + 1/2at^2

d1 = 1/2gt^2

d1 = 1/2(9.8)(6)

d1 = **29.4m.**

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