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

2seconds

Explanation:

according to Newton second law

s=ut+1/2 at*t

here u( initial velocity) =0

therfore ut=0

so s=1/2 gt*t. (as a = g= 10 m/sec^2)

s= 1/2 gt* t

so

t = (2s/g)^1/2

t =(40/10)^1/2= 2sec

Answer 2

Answer:-

2 seconds

Explanation:-

s=ut+1/2 at×t

twe know that u(initial velocity)=0

twe know that u(initial velocity)=0therefore,ut=0

twe know that u(initial velocity)=0therefore,ut=0so, s=1/2gt×t (as a=g=10m/sec^2)

s=1/2gt×t

so,

t=(2s/g)^1/2

t=(40/10)^1/2

=2 seconds

Hope this will help you ☺️☺️