Question

A boy dropped a ball from the top of 20 m
building and starts the stopwatch. What will be
the reading of stopwatch when ball will reach
at ground surface, if acceleration due to gravity
is 10 m/s ^2?

Answer 1

Answer:

2 seconds

Explanation:

Given:

• Height of the building = s = 20 metres
• Acceleration due to gravity = g = 10 m/s²
• Initial velocity = u = 0m/s (The ball will start initially from rest)

To find:

• Time taken for the ball to reach the surface of the earth

For finding the time in this question, we can use the second equation of motion which says:

$S=ut+\frac{1}{2} at^{2}$

Substituting the values:

$20=0 \times t + \frac{1}{2} \times 10 \times t^{2}$

$20= 5 \times t^{2}$

$20= 5 t^{2}$

$t^{2} =\frac{20}{5}$

$t^{2} =4$

$t=\sqrt{4}$

$t=2$ seconds

The time taken for the ball to reach the ground is equal to 2 seconds

Answer 2

Answer:

Explanation:

Given :-

Height of building, s = 20 m

Acceleration due to gravity, a = 10 m/s²

Initial velocity, u = 0 (As ball starts from rest)

To Find :-

Time taken by ball to reach on earth's surface

Formula to be used :-

2nd equation of motion

i.e s = ut + 1/2 × at²

Solution :-

Putting all the values, we get

s = ut + 1/2 × at²

⇒ 20 = 0 × t + 1/2 × 10 × t²

⇒ 20 = 5 × t²

⇒ 20/5 = t²

⇒ t² = 4

⇒ t = √4

⇒ t = 2 seconds

Hence, the time taken by ball to reach on earth's surface is 2 seconds.