Question

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground?

NCERT Class IX
Sciences - Main Course Book

Chapter 8. Motion

Answer 1

Given that,
Distance(s)="20m"
Initial velocity(u)=0m/s
Accelaration="10m/s-2"

Using the relation Distance-Time

s=v2 - u2=2a

s= v2-0=2*10
20=v2-0=20
v2-0=20*20
v2=400(underoot)
v=20 m/s

time=v=u=at
20=0+10
20 upon 10=0+10(t)=
time= 2 secs

Answer 2

Answer :-

Assume, the final velocity with which the ball will strike the ground be ‘v’and time it takes to strike the ground be ‘t;

Initial Velocity of ball, u =0

Distance or height of fall, s =20 m

Downward acceleration, a =10 m s-2

As we know, 2as =v2-u2

v2 = 2as+ u2

= 2 x 10 x 20 + 0

= 400

∴  Final velocity of ball, v = 20 ms-1

t = (v-u)/a

∴Time taken by the ball to strike = (20-0)/10

= 20/10

= 2 seconds