A rock is shot vertically upward from the edge of the top of a tall building. The rock reaches its maximum height above the top of the building 1.60 s after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground 7.00 s after it is launched. In SI units: (a) with what upward velocity is the rock shot, (b) what maximum height above the top of the building is reached by the rock, and (c) how tall is the building?

Question

Physics

Kyler Wilkerson

17 June, 03:56

A rock is shot vertically upward from the edge of the top of a tall building. The rock reaches its maximum height above the top of the building 1.60 s after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground 7.00 s after it is launched. In SI units: (a) with what upward velocity is the rock shot, (b) what maximum height above the top of the building is reached by the rock, and (c) how tall is the building?

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Answers (1)

Know the Answer?

Collier · Answer 1

a) 15.68 m/s

b) 130.34 m

Explanation:

Let's consider the origin of the coordinates at the ground, therefore, the equation of motion of the rock and its velocity are

x (t) = h + vt - (1/2) gt^2

v (t) = v - gt

Where h is the height of the building and v is the initial velocity.

The maximum height is reached when v=0, that is v (1.6s) = 0, and we know that x (7s) = 0

Therefore

0 = v (1.6s) = v - (9.8 m/s^2) (1.6s)

v = 15.68 m/s

and

0 = x (7s) = h + (15.68 m/s) (7 s) - (1/2) (9.8 m/s^2) (49s^2)

h = (1/2) (9.8 m/s^2) (49s^2) - (15.68 m/s) (7 s)

h = 130.34 m