Question

A ball of mass 2 kg is dropped from the top of a tower 45 m high. Find the time required by it to fall to the ground and the velocity with which it hits the ground

Answer 1

Given :-

Mass of ball = 2 kg

Height of the tower, H = 45 m

Initial velocity, u = 0 m/s

To Find :-

Time required to fall to the ground (t).

Velocity of the ball when it hits the ground (v).

Solution :-

We can find time required to fall to the ground by using the third equation of kinematics.

$\boxed{\bf H = ut+\dfrac{1}{2} gt^2}$

$\longrightarrow \sf 45 = 0 \times t+ \dfrac{1}{2} \times 10 \times t^2 \\\\\longrightarrow 45= 0+\dfrac{10t^2}{2} \\\\\longrightarrow 10t^2= 45 \times 2 \\\\\longrightarrow 10t^2 = 90 \\\\\longrightarrow t^2 = 9\\\\\longrightarrow \bf t = 3 \ seconds$

Time required to fall to the ground = 3 seconds

We can find the velocity of the ball when it hits the ground by using the first equation of kinematics.

$\boxed{\bf v = u+gt}$

$\longrightarrow \sf v = 0 + 10 \times 3 \\\\\longrightarrow \bf v = 30 \ m/s$

Velocity of the ball when it hits the ground = 30 m/s