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/s2?

Answer 1

Answer:

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

Answer 2

Answer:

The reading on the stopwatch will be 2 seconds.

Explanation:

The second equation of motion for a freely falling body says that

$s = \frac{1}{2} g {t}^{2}$

So,

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

or

${t}^{2} = 4$

or

$t = 2s$

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