Question

A body is projected horizontally with speed v metre per second from height h it takes 8 seconds to reach the ground the time it takes to cover the first one fourth of height is?

Answer 1

Given:

A body is projected horizontally with speed v metre per second from height h it takes 8 seconds to reach the ground.

To find:

Time taken to cover the first ¼th of height.

Calculation:

This is a case of Height to Ground Projectile.

The initial Y component of velocity is zero.

$\therefore \: h = ut + \dfrac{1}{2} g {t}^{2}$

$= > \: h = (0 \times t) + \dfrac{1}{2} g {t}^{2}$

$= > \: h = \dfrac{1}{2} g {t}^{2}$

$= > \: h = \dfrac{1}{2} \times 10 \times {(8)}^{2}$

$= > \: h = 320 \: m$

So , ¼th of height shall mean 80 m.

Again applying the same equation:

$\therefore \: h2 = u(t2) + \dfrac{1}{2} g {(t2)}^{2}$

$= > \: h2 = \dfrac{1}{2} g {(t2)}^{2}$

$= > \: 80= \dfrac{1}{2} \times 10 \times {(t2)}^{2}$

$= > {(t2)}^{2} = 16$

$= > t2 = 4 \: sec$

So , time taken to complete the first ¼th height is 4 seconds.

Answer 2

Answer: 4s

Explanation: t = √2h/g ( Horizontal

projection)

Then, t²1 = 2h/g ______(1)

t²2 = 4×2h/g ______(2)

(2) ÷ (1)

t²2/(8)² = 1/4

t2 = √64/4 =√16 = 4sec