Question

A child drops a ball from a height of 10m. Assume that its

velocity increases uniformly at the rate of 10 m/s2

,find:

i. The velocity with which ball strikes the ground and

ii. The time taken by the ball to reach the ground.



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Answer 1

Answer:

Explanation:

Given,

Height attained, s = 10 m

Acceleration, a = 10 m/s²

To Find,

1. Final velocity, v

2. Time taken, t

Formula to be used,

1st equation of motion, i.e v = u + at

3r equation of motion, i.e v² - u² = 2as

Solution,

1. Final velocity

Putting all the values, we get

v² - u² = 2as

⇒ v² = 2as (As u = 0 m/s)

⇒ v² = 2 × 10 × 10

⇒ v² = 200

⇒ v = √200

⇒ v = 10√2 m/s

Hence, the final velocity is 10√2 m/s.

2. Time taken,

v = u + at

⇒ 10√2 = 10 × t

⇒ 10√2 = 10t

⇒ 10√2/10 = t

⇒ t = √2 seconds

Hence, the time taken is √2 seconds.

Answer 2

Given :-

A child drops a ball from a height of 10m. Assume that its  velocity increases uniformly at the rate of 10 m/s²

To Find :-

Velocity

Time taken

Solution :-

We know that

v² - u² = 2as

v² - 0 = 2(10)(10)

v² - 0 = 200

v² = 200 + 0

v² = 200

v = √200

v = 10√2 m/s

Now

a = v - u/t

t = v - u/a

t = 10√2 - 0/10

t = √2/1

t = √2 sec