Question

A ball is gently dropped from a height of 20m. If its velocity increases uniformly

at the rate of 10ms-2, with what velocity will it strike the ground? After what

time will it strike the ground?​

Answer 1

Given:-

→ Height from which the ball is dropped = 20m

→ Acceleration of the ball = 10 m/s²

To find:-

→ The velocity with which the ball will

strike the ground.

→ Time after which the ball with strike the

ground.

Solution:-

In this case :-

• Initial velocity (u) of the body will be

zero as it was initially at rest.

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By using the 3rd equation of motion, we get :-

=> v² = u² + 2as

=> v² = 0 + 2(10)(20)

=> v² = 400

=> v = √400

=> v = 20 m/s

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Now, from the 1st equation of motion :-

=> v = u + at

=> 20 = 0 + 10(t)

=> 20 = 10t

=> t = 20/10

=> t = 2s

Thus :-

• The ball strikes the ground with a

velocity of 20m/s .

• Also, it strikes the ground after 2s .

Some Extra Information:-

The 3 equations for a body moving with uniform acceleration are :-

→ v = u + at

→ s = ut + ½at²

→ v² - u² = 2as

Answer 2

We have,

Initial height = 20 m

Acceleration = + 10 m/s²

Initial velocity = 0 m/s

We know,

2as = v² - u² (third equation of linear kinematics)

⇒ 2gh = v²

⇒ v = √2gh = √(2 × 10 m/s² × 20 m)

⇒ v = √(20 × 20 m²/s²)

⇒ v = 20 m/s ...(1)

Now we know,

v = u + at (first equation of linear kinematics)

⇒ v = gt

⇒ t = v/g = (20 m/s)/(10 m/s²) = 2 s ...(2).