Question

a boy standing at the top of a tower of 20m height drops a stone. assuming g= 10m/s², the velocity with which it hits the ground is?

Answer 1

h=20m
g=10m/s^2
u=0
v^2-u^2 = 2gh
v^2-0=2*10*20
v = 20 m/s^2

Answer 2

Answer:

The velocity is 20 m/s

Given:

Height of the tower, H = 20 m

Initial velocity, u = 0 (as drops)

Final velocity, v = ?

Solution:

The boy standing at top of a tower drops a stone from the tower height of 20 m, velocity with which stone will hit ground will be calculated as follows,

From the equation of motion, we have,

$\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{a} \mathrm{s}$

$\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gH}$

$\mathrm{v}^{2}=(0 \times 0)+(2 \times 10 \times 20)$

$\mathrm{v}^{2}=200$

$\mathrm{v}=20 \ \mathrm{m} / \mathrm{s}$

The stone touches the ground with the velocity of 20 m/ s when dropped from a height of 20 m.