Question

A stone is released from the top of a tower of height 19.6m.Its final velocity just before touching the ground.

Answer 1

Hi, my friend,

Given that:

Ball is release which means initial velocity of the ball (u) = 0 m/s

Height of the tower = 19.6 m

g = 9.8 m/s² (Consider down as positive)

Formula to be used:

v² - u² = 2gh

=> v² - (0)² = 2(9.8)(19.6)

=> v² = 2(9.8)(19.6)

=> v² = 384.16 m²/s²

=> v = + or - √(384.16) m/s

=> v = + or - 19.6 m/s

[As we considered downwards as positive]

=> v = + 19.6 m/s

Hence the final velocity just before the ball touches the ground in 19.6 m/s.

Harith

Maths Aryabhatta

Physics Genius.

Answer 2

a = 9.8 m s^-2

final velocity = u +(a *t)

u =0
a =9.8
t= 19.6/9.8 =2s

v= u + at
=0 + 9.8 m s^-2 *2s
= 19.6 m s^-1

answer:

v = 19.6 m/s