Question

A ball is gently dropped from a height 30m. If its velocity increases uniformly at the rate of 10m/s^2, with what velocity will it strike the ground and at what time?

Answer 1

Answer:-

v = 24.50 m/s

t = 2.45 s

Given :-

$h = 30 m \\ u = 0 m/s \\ g = 10 m/s^2$

To find :-

The strike velocity and time.

Solution:-

Let the strike velocity be v and time be t.

• By using 3rd equation of motion.

→$2gh = v^2 - u^2$

• Put the given values,

→$2 \times 10 \times 30 = v^2 - (0) ^2$

→$600 = v^2$

→$v= \sqrt{600}$

→$v = 24.50 m/s$

hence,

Velocity of strike is 24.50 m/s.

• Using first equation of motion.

→$v = u+gt$

→$24.50 = 0 + 10 \times t$

→$24.50 = 10t$

→$t = \dfrac{24.50}{10}$

→$t = \dfrac{2450}{1000}$

→$t = 2.45s$

hence,

The time taken by ball to reach the ground is 2.45s.