Question

A stone is released from the top of tower of height 19.6m. Calculate it's final velocity. Just before touching the ground.


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Answer 1

Given:-

•Height of the tower(h)=19.6m

•Acceleration due to gravity(g)=9.8m/s²

•Initial velocity(u)=0 (as the stone was at rest initially)

To find:-

•Final velocity or velocity of the stone before touching the ground(v)

Solution:-

By using the 3rd equation of motion,when u=0,we get-----

=>v²=2gh

=>v²=2×9.8×19.6

=>v²=384.16

$= > v = \sqrt{384.16}$

=>v=19.6m/s

Thus,velocity of the stone before touching the ground is 19.6m/s.

Answer 2

Answer:

I hope this help you

Explanation:

initial velocity u=0

Acceleration,a=g=9.8ms-2

Distance traveled,s=19.6m

let the final velocity be v

According to the third equation of motion

u2-u2=2as

u2-02=2×9.8×19.6u=√2×9.8×19.6

u=19.6ms-1

its just final velocity before tuouching the ground will be=19.6ms-1

Thankyou........