Question

A rock is dropped from the top of a tall building of height 70m the speed of the rock as it strikes the ground

Answer 1

u=0

v=?

g=10ms⁻¹

h=70m

v²=u²+2gh

v²=2*10*70

v=√1400

v=10√14ms⁻¹

Answer 2

Given:

• Rock is dropped from the top of a tall building of height 70 metres.

• Acceleration due to gravity = 10 m/s².

To find:

• Velocity with which the rock strikes the ground ?

Calculation:

• Since the rock is dropped from the top of the building, the initial velocity will be zero.

• Also, the rock will be experiencing a constant gravitational acceleration of g = 10 m/s².

Applying EQUATIONS OF KINEMATICS:

${v}^{2} = {u}^{2} + 2as$

$\implies \: {v}^{2} = {u}^{2} + 2gh$

$\implies \: {v}^{2} = {(0)}^{2} + 2(10)(70)$

$\implies \: {v}^{2} = 1400$

$\implies \: v = \sqrt{1400}$

$\implies \: v = 37.41 \: m {s}^{ - 1}$

So, velocity with which the stone will strike the ground is 37.41 m/s