Question

7. A ball is gently dropped from a height of 20 m. If its velocity
increases uniformly at the rate of 10 m s?, with what velocity
will it strike the ground? After what time will it strike the
ground?
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Answer 1

Hey mate,.

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

Hope it will help you

✌️sai

Answer 2

Here is your answer.....

Answer.....

Given: height=20m

It's Velocity increases uniformly at the rate of 10 m s. It's simply acceleration (due to gravity).

Initial velocity=u=0

Acceleration due to gravity=g=10 m s...

To find: The velocity of the ball as it strikes the ground...

Soln....

Final velocity v= u+at where t is the time which takes to hit the ground..

Using 2 equation of motion...

h=ut 1/2×gt×gt

20=0×t 1/2 x10 ×10×t×t

20=0 ×5 ×t×t

20/5=t2

4=t2

±2=t

so, time cannot be negative hence time will be 2 second.

Now using first equation of motion....

v=u+gt

v=0×10 ×2

v=20

So velocity it takes to hit the ground is 20 m/s...

hope it will help you....