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

Answer:

2 seconds

Explanation:

Here u = initial velocity = 0 m/s

        v - final velocity = ?

        t = total time = ?

        a = acceleration = 10 m/s

using equation of motion

2ax = v^2 - u^2

2(10)(20) = v^2 - 0

400 = v^2

v = 20 m/s

Again using equation of motion

v = u + at

20 = 0 + 10t

20/10 = t

t = 2s

Answer 2

Answer:-

10√5 m/s

Solution:-

Height(s)→20m

Velocity(u)→ 10m/s

It is dropping gently.

To Find:- Velocity(v) at which bit will strike

Assuming downward as +ve and g→10m/s

Thus, we will use here equation of motions.

Now,

Finding Velocity:-

From Equation of motion we have,

v²=u²+2as

→v²= (10)²+2(10)(20)

→v²=100+400

→v=10√5 Answer

Hence, our answer is 10√5 m/s.

Hope It helps You.