Question

A ball is gently dropped from a height of 20m if its velocity increases uniformly at the rate of 10 metre per second square with what velocity will it strike the ground? After what time will it strike the ground?​

Answer 1

given,

distance (s)= 20m

acceleration (a) = 10m/s^2

initial velocity (u)=0m/s

final velocity (v) =?

time ( t )=?

we know ,

v square - u square =2as

v square -0=2×10×20

v square =400

v=√400

v= 20m/s

and,

v=u+at

20=0+10×a

20=10t

t=2 sec

striking velocity 20 metre per second .

striking time 2 second .

hope it help u ✔

Answer 2

$\huge \boxed{ \bf \green{answer}}$

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