Question

A stone drops from the edge
of a roof. It passes a windows 2 metres high in 0.1 second.How far is the roof above the top of the window.​

Answer 1

Given that, the height (h or s) of the window is 2 m and time (t) is 0.1 sec.

We have to find how far (s) is the roof above the top of the window.

Using the Second Equation of Motion,

s = ut + 1/2at²

2 = u(0.1) + 1/2 × 10 × (0.1)²

2 = 0.1u + 5(0.01)

2 = 0.1u + 0.05

2 - 0.05 = 0.1u

1.95 = 0.1u

u = 19.5 m/s

Therefore, the velocity of the stone is 19.5 m/s when it covers a distance between roof and window.

But the initial velocity of the stone is 0 m/s when it is at rest.

Now, using the Third Equation of motion,

v² - u² = 2as

(19.5)² - (0)² = 2(10)s

380.25 - 0 = 20s

s = 380.25/20

s = 19.0125

s = 19 (approx.)

Therefore, the height of the roof above the top of the window is 19 m.

Answer 2

Correct Question -

A stone drops from the edge of a roof.

It passes a windows 2 metres high in 0.1 second.

How far is the roof above the top of the window ?

Solution -

In the above Question, the following information is given -

A stone drops from the edge of a roof.

It passes a windows 2 metres high in 0.1 second.

We have to find how far the roof is above the top of the window .

Here, the height of the window is 2 m.

Time, t = 0.1 second.

The stone is dropped from the edge of the roof .

Hence, it's innitial velocity is 0

Let the final Velocity of the stone be V

Now,

According to Second Equation Of Motion ,

S = Vt + ( 1 / 2 ) a t ^ 2

Here, S is e Vertical displacement , H

Therefore H = 0.1 V + ( 1 / 2 ) × ( -g ) × 0.01

=> 2 = 0.1 V + 0.05

=> V = 19.5 m / s.

Now, according to the third Equation of motion -

V^ 2 - U ^ 2 = 2AS

Here,

V ^ 2 = 2g D [ As U = 0, a = -g , D = Distance between Roof and Window ]

=> ( 19.5 ) ^ 2 = -20 D

=> D = 19 metre approximately.

Hence the roof is 19 m above the top of the window approximately.

Answer -

The roof is 19 m above the top of the window approximately