Question

A stone drops from the top of a cliff, from a height of 19.6 metres above the ground.a)Find the speed of the stone after 0.5 seconds.b)Find the speed of the stone when it hits the ground.C)Find the time taken for the stone to hit the ground.

Answer 1

Given:

The height of the cliff, h=19.6 m

The initial velocity of the stone, u=0 m/s

The time period, t₀=0.5 s

To find:

a) The speed of the stone after 0.5 s

b) The speed of the stone when it hits the ground.

c) The time it takes for the stone to hit the ground.

Explanation:

a)

From the equation of motion,

$v_0=u+gt_0$

Where g is the acceleration due to gravity and v₀ is the velocity of th stone after 0.5 s

On substituting the known values,

$\begin{gathered} v_0=0+9.8\times0.5 \\ =4.9\text{ m/s} \end{gathered}$

b)

From the equation of motion,

$v^2-u^2=2gh$

Where v is the velocity of the stone when it hits the ground.

On substituting the known values,

$\begin{gathered} v^2-0=2\times9.8\times19.6 \\ \implies v=\sqrt{2\times9.8\times19.6} \\ =19.6\text{ m/s} \end{gathered}$

c)

From the equation of motion,

$h=ut+\frac{1}{2}gt^2$

Where t is the time it takes for the stone to reach the ground.

On substituting the known values,

$\begin{gathered} 19.6=0+\frac{1}{2}\times9.8\times t^2 \\ \implies t=\sqrt{\frac{19.6\times2}{9.8}} \\ =2\text{ s} \end{gathered}$

Final answer:

a) The speed of the stone after 0.5 s is 4.9 m/s

b) The speed of the stone when it hits the ground is 19.6 m/s

c) The time it takes for the stone to reach the ground is 2 s