Question

8) A body is released from the top of a building of height 19.6 m, find the velocity with w
hits the ground
1) 9.6 m/s
2) 0.96 m/s 3) 19.6 m/s 4) 19.6 m/s2​

Answer 1

Answer:

3) 19.6 m/s

Explanation:

u=o v=?

v

$v { }^{2} = u {}^{2} + 2as$

s=19.6m , a=9.8m/s2

$v = \sqrt{2as}$

v=19.6m/s

Answer 2

The velocity of the body when hits the ground is 19.6 m.

Explanation:

Given height of the building, s = 19.6 m.

To calculate the velocity of the body when it hits the ground use third equation of motion as

$v^2=u^2+2gs$

Here, u is initial velocity, v is the final velocity and g is acceleration due to gravity.

As body released from rest, so u = 0.

Substitute the given values, we get

$v^2=0+2\times9.8m/s^2\times19.6m$

$v=\sqrt{2\times 9.8m/s^2\times19.6m}$

v = 19.6 m/s.

Thus, the velocity of the body when hits the ground is 19.6 m.

#Learn More: Equation of motion.

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