Question

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

Answer 1

Answer:

Step-by-step explanation:  t = √2h/g

                    then putting the values give t= 2 sec

Answer 2

Answer:

Initial velocity = 20 $\frac{m}{s}$

Final velocity = 0 $\frac{m}{s}$ (it stops at the ground)

Acceleration = - 10 $\frac{m}{s^{2} }$

Therefore, time = $\frac{v-u}{a}$ = $\frac{0 - 20}{- 10}$ = $\frac{-20}{-10}$ = 2 s

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