Question

A stone is thrown horizontally from the top of a vertical cliff. Given that the initial
velocity of the stone is 20 m/s and that it hits the horizontal ground below the cliff
3 seconds later, calculate: (1) the horizontal distance travelled by the stone from
the foot of the cliff. (2) The height of the cliff.

Answer 1

Answer :

The horizontal distance travelled by the stone from  the foot of the cliff = 60 m

The height of the cliff = 44.1 m

Explanation :

Given :

• The initial  velocity of the stone is 20 m/s
• It hits the horizontal ground below the cliff  3 seconds later.

To find :

• the horizontal distance travelled by the stone from  the foot of the cliff
• the height of the cliff

Solution :

   The horizontal distance covered by projectile during the time of flight is given by,

     $\boxed{\sf R=uT}$

where

➙ u denotes initial velocity

➙ T denotes time of flight

Substitute the values,

R = 20 m/s × 3 s

R = 60 m

∴ The horizontal distance travelled by the stone from  the foot of the cliff is 60 m

 The time of flight is given by,

     $\boxed{\sf T=\sqrt{\dfrac{2h}{g}}}$

where

➙ h denotes height of the building/cliff/tower

➙ g denotes acceleration due to gravity

Substitute the values,

 $\sf 3=\sqrt{\dfrac{2h}{9.8}} \\\\ \sf 3^2=\dfrac{2h}{9.8} \\\\ \sf 9=\dfrac{2h}{9.8} \\\\ \sf 2h=9 \times 9.8 \\\\ \sf 2h=88.2 \\\\ \sf h=88.2/2 \\\\ \sf h=44.1 \ m$

∴ The height of the cliff is 44.1 m