Question

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate
of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the
ground?

Answer 1

GIVEN -

Height (h) = 20 m

Acceleration due to gravity(g) = 10 m/s²

When we drop a ball, the ball has initial velocity (u) as 0 m/s

TO FIND -

The final velocity

The time taken

SOLUTION-

The third equation of motion -

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

where  v = final velocity, u = initial velocity, g = acceleration due to gravity and s = height

Substituting the values of u, g and s,

$v^{2} = 0^{2} + 2(10)(20)\\\\v^{2} = 400 \\\\v= \sqrt{400} = 20 m/s$

Thus, the velocity with which it will strike the ground is 20 m/s

The first equation of motion,

$v = u+gt$

where v = final velocity, u = initial velocity, g = acceleration due to gravity and t = time

Substituting the values of v, u and g

$20 = 0 + 10(t)\\\\10t = 20\\\\t = \frac{20}{10} = 2 s$

After 2 seconds, the  ball will strike the ground.

ANSWER -

The velocity with which it will strike the ground is 20 m/s.

After 2 seconds, the  ball will strike the ground.