Question

A child drops a ball from a height of 10m. Assume that its velocity increases
uniformly at the rate of 10m/s2 . Find (i) the velocity with which the ball strikes the ground (ii) the time taken by the ball to reach the ground.

Answer 1

Given:-

• Height (s) = 10 m
• Velocity increases means acceleration (a) = 10 m/s²

To find:-

• i) Final velocity with which the ball hits the ground.
• ii) Time taken by the ball to reach ground.

Solution:-

i) Here, given, s = 10m, a = 10 m/s²

And the ball has been dropped freely,so here, initial velocity (u) = 0

As we have to find final velocity (v) so we will use this equation of motion.

☛ v² = u² + 2as

→ v² = 0² + 2 × 10 × 10

→ v² = 0 + 200

→ v² = 200

→ v = √200

→ v = √(100 × 2)

→ v = √(10² × 2)

→ v = 10√2 m/s

Therefore,

Velocity of ball while striking the ground is (10√2 m/s²)

$\rule{200}{1}$

ii) Now we got values of v = 10√2 m/s ,u = 0 , a = 10 m/s²

As here,we have to find out time to reach the ground,we will use this equation of motion:-

☛ v = u + at

→ 10√2 = 0 + 10t

→ 10√2 = 10t

→ t = (10√2)/10

→ t = √2

→ t = 1.41 s

Therefore,

Time taken by the ball to reach ground is (1.41 s) .

Answer 2

Solution :

⏭ Given:

✏ Initial velocity of ball = 0 mps

✏ Height = 10 m

✏ Gravitational acceleration = 10 m/s$^2$

⏭ To Find:

• The velocity with which the ball strikes the ground
• The time taken by ball to reach the ground

⏭ Formula:

✏ As per second equation of kinematics

$\bigstar\:\boxed{\sf{\pink{\large{v^2-u^2=2as}}}}$

✏ As per third equation of kinematics

$\bigstar\:\boxed{\sf{\purple{\large{s=ut+\dfrac{1}{2}at^2}}}}$

⏭ Calculation:

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• Calculation of final velocity

$\mapsto\sf\:v^2-(0)^2=2as \\ \\ \mapsto\sf\:v=\sqrt{2gs} \\ \\ \mapsto\sf\:v = \sqrt{2\times10\times10}\\ \\ \mapsto\sf\:v = \sqrt{200}\\ \\ \mapsto\: \boxed{\sf{v = 10\sqrt{2}\: mps}}$

✒ The ball strikes the ground with velocity 10√2 mps

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• Calculation of time

$\leadsto\sf\:s=(0)t+\dfrac{1}{2}at^2 \\ \\ \leadsto\sf\: s=\dfrac{1}{2}at^2 \\ \\ \leadsto\sf\: t=\sqrt{\dfrac{2s}{g}}\\ \\ \leadsto\sf\:t=\sqrt{\dfrac{2\times10}{10}}\\ \\ \leadsto\: \boxed{\sf{t=\sqrt{2}\: s}}$

✒ Time taken by ball to reach the ground = √2sec

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