Question

1. A stone is dropped from a height of 10 m above
the top of a window of 4.4 m high (top to bottom of
the window). Find the time taken by the stone to
cross from the top to bottom of the window.​

Answer 1

Given:-

• Initial velocity of Stone = 0m/s

• Height above the top of a window = 10m

• Height of window = 4.4

• Acceleration due to gravity = + 10m/s²

To Find:-

• The Time taken by the stone to cross the top to bottom of the window.

Formulae used:-

• s = ut + ½ × a × t²

Where,

• s = Distance
• u = Initial Velocity
• a = Acceleration
• t = Time

Now,

Time taken to cross the height of 10m

→ s = ut + ½ × a × t²

→ 10 = 0 × t + ½ × 9.8 × t²

→ 10 = 4.9t²

→ t² = 10/4.9

→ t² = 2.04

→ t (x) = 1.42s

Hence, The time taken to cross the 10m height is 1.42s

Again,

→ Total Distance = 10 + 4.4

→ Total Distance = 14.4m

→ s = ut + ½ × a × t²

→ 14.4 = 0 × t + ½ × 9.8 × t²

→ 14.4 = 0 + 4.9t²

→ 14.4 = 4.9t²

→ t² = 14.4/4.9

→ √t² = √2.93

→ t ( y ) = 1.71s

Hence, The time taken by stone to cross whole height is 1.7s

Therefore,

→ Time taken to cross the top of window = y - x

→ 1.71 - 1.42

→ 0.29s

Therefore, The time taken by stone to

stone tocross from the top to bottom of the wwindo is 0.29s.

Answer 2

Given:-

• A stone is dropped from a height of 10 m above

• the top of a window of 4.4 m high (top to bottom of the window).

• Initial velocity of Stone = 0m/s

To Find: -

• The Time taken by the stone to cross the top to bottom of the window.

Formulae used:-

$\boxed{ \sf \: s = ut + \frac{1}{2}g {t}^{2} }$

Substitute all values :

$\sf : \implies \: 14.4 = 0 \times t \: + \frac{1}{2} \times 10 \times {t}^{2} \\ \\ \sf : \implies \: \: 14.4 = \cancel{\frac{10}{2} }{t}^{2} \\ \\ \sf : \implies \: \: 14.4 = 5 {t}^{2} \\ \\ \sf : \implies \: {t}^{2} = \frac{14.4}{5} \\ \\ \sf : \implies \: {t}^{2} = 2.88 \\ \\ \sf : \implies \:t = 1.71s$