Question

A ball is gently dropped from a height of 20m.if its velocity increases uniformly at the rate of 10m/s with what velocity will it strike the ground ?after what time will it strike the ground

Answer 1

GIVEN:

s=20m

a=10m/sec

u=0m/sec

TO FIND:

v=?

t=?

SOL :

By 3rd equation of motion i.e.

2as= v square - u square

2 multiply by 10 multiply by 20 = v square-0square

400= v square

v = square root of 400

v = 20m /sec

It will strike the ground with a velocity of 20m/sec.

By 1st equation of motion

v=u+at

20 = 0 + 10 multiply by t

20=10t

t=20/10

t = 2 sec

It will strike the ground after 2 secs.

Answer 2

given a=10m/s²
s=20 m
u=0m/s

now using s=ut+½at²
or 20=½*10t². (since u=0)
or t=2 sec

for final velocity
now v=u+at
v=0+2*10
20m/s