Question

A stone is dropped from the top of a tower 45 m high. What is its velocity when it hits the ground? (Given g = 10 ms–2).

Answer 1

Explanation:

By third law of motion

s=ut+1/2gt²

45=1/2

Answer 2

Answer:

velocity when it hits the ground = 30 m/s

Explanation:

Given,

Height of tower ,h = 45 m

acceleration due to gravity,$g\ =\ 10\ m /s^2$

initial velocity of stone, u = 0

final velocity, v = ?

from the equation of motion, we have

$v^2\ =\ u^2\ +\ 2.g.h$

$=>\ v^2\ =\ 0\ +\ 2\times 10\times 45$

$=>\ v^2\ =\ 900$

$=>\ v\ =\ 30\ m/s$

So, the velocity of stone when it hits the ground will be 30 m/s.