Question

A particle is dropped from the top of a tower of height 4.9 m the velocity of particle with which it strikes the ground is

Answer 1

We will use the following equation of motion.

V² = U² + 2gs

In this case :

V = Final velocity

U = initial velocity

g = gravitational acceleration

S = height of the tower.

The initial velocity equals to 0

Doing the substitution we have :

V² = 0 + 2 × 9.8 × 4.9 = 96.04

V² = 96.04

V = 9.8

= 9.8m/s

Answer 2

Given height of the tower h = 4.9 m.

Initial velocity u = 0.

g = 9.8 m/s^2.

We know that v^2 = u^2 + 2gh

= > v^2 = 2 * 9.8 * 4.9

= > v^2 = 96.04

= > v = 9.8.

Therefore, The velocity of the particle = 9.8 m/s.

Hope this helps!