You’re standing on the roof of your sci- ence building. You drop a rock from res

Question

You’re standing on the roof of your sci- ence building. You drop a rock from rest and notice that it takes an amount of time T to hit the ground. Express your answers to the follow- ing questions in terms of T and the acceleration due to gravity, g.

(a) How high is the building?

(b) How fast must you throw the rock straight up

if the rock is to take 2T to hit the ground?

(c) For the situation described in part b, how

long does it take from the time you let the rock go

to when the rock reaches maximum height?

(d) For the same situation, what is the maximum

height that the rock gets to above the ground?

Explanation / Answer

(a)

You drop a rock from rest.

so, u = 0

Let, height of building = H

From second kinematic equation,

s = ut + (1/2)at^2

H = (1/2)gT^2

(b)

Consider the downward direction = +ve

Here, Total displacement of rock = H

s = ut + (1/2)at^2

H = -u*2T + (1/2)g*(2T)^2

Put the value of H from part a,

(1/2)gT^2 = 2gT^2 - 2uT

u = 3gT / 4

(c)

At maximum height, v = 0

v = u + at

0 = 3Tg / 4 - gt

t = 3T / 4

(d)

let the height above the building = h

v^2 = u^2 + 2as

0 = (-3gT / 4)^2 + 2gh

h = -9gT^2 / 32

Maximum height that the rock gets to above the ground,

Hmax = H + h

Hmax = (1/2)gT^2 - 9gT^2 / 32

Hmax = 7gT^2 / 32

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