Question

1. A body is projected horizontally with speed 5 m/s
from height h, takes 8 s to reach the ground. The
time it takes to cover the first one fourth of height
is
[NCERT Pg. 79]
(1) 4 s
(2) 6 S
(3) 2 s
(4) 5s
why initial velocity taken to be zero in fact 5m/s velocity already given in question ​

Answer 1

Answer:

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=ut+

2

1

gt

2

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

2

1

gt

2

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

2

1

gt

2

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

2

1

×10×(8)

2

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

So , ¼th of height shall mean 80 m.

Again applying the same equation:

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

2

1

g(t2)

2

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

2

1

g(t2)

2

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

2

1

×10×(t2)

2

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

2

=16

= > t2 = 4 \: sec=>t2=4sec

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