Question

1. A body is dropped from a height of 80 m. Find the time taken by the body to reach the ground

and velocity on reaching the ground. Take g = 10 m/s.​

Answer 1

Answer:

Time take=4 seconds

Velocity on reaching ground =40 m/s

Explanation:

s=ut+0.5at²

80=(0×t)+(0.5×10×t²)

t²=16

t=4s

v=u+at

v=0+(10×4)

v=40 m/s

Answer 2

Answer:

The time taken by the body to reach the ground is equal to 4sec and the velocity of body on reaching the ground will be equal to 40m/s.

Explanation:

Given,

The body covered the distance before reaching the ground, $S=80m$

Consider that u is the initial velocity of the body,

Because the body was at rest initially  ∴  u=0

Consider that the v is the final velocity of body on reaching the ground.

And, t is time taken by body.

The body will fall under the act of gravity, so the acceleration will be gravitational acceleration.

$a=g=10ms^{-1}$

From the 2nd equation of motion:

$S=ut+\frac{1}{2}at^{2}$

$S=ut+\frac{1}{2}gt^{2}$

$80=(0)t+\frac{1}{2} (10)t^{2}$

$80=5t^{2}$

$t^{2}=16$

$t=4sec$

From the first equation of motion:

$v=u+at$

$v=u+gt$

Put the value of u, g and t in above equation.

$v=(0)+(10)(4)$

$v=40ms^{-1}$

Therefore, the time taken by the body is equal to 4sec and final velocity of the body is 40m/s.