Question

A ball is gently dropped from a height of 20 CM. If its velocity increases uniformly at the rate of 10 M per second square. with what velocity will it strike the ground? After what time will it strike the ground?

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:

20 m/s, 2 s

Explanation:

Given, initial velocity of ball, u=0

Final velocity of ball, v=?

Distance through which the balls falls, s=20m

Acceleration a=10ms−2

Time of fall, t=?

We know

v^2−u^2=2as

or v^2−0=2×10×20=400

v=√400

v=20ms−1

Now using

 v=u+at 

we have

20=0+10×t 

20=10t

t=20÷10

t=2s