Question

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

Answer 1

Here, u= 0, v= ?, a= 10, s= 20, t= ?

As we know,
v² = u²+2as
=> v² = (0)^2+ 2(10)(20)
=> v² = 400
=> v = 20
Now,
s = ut + 1/2at²
20 = 1/2×10×t²
t² = 20/5
t² = 4
t = 2

I hope this will help you.

please select it as the brainliest answer.

Answer 2

Initial velocity(u) = 0 m/s
Acceleration (a) = 10 m/s²
Time(s) = 20 m

We know,

v² = u² + 2as

Putting values:

v² = 0² + 2 (10 x 20)
v² = 400
v = 20 m/s

For time:
 v = u + at
t = v - u/a
t = (20-0)10
  = 20/10
  = 2

∴ Striking velocity = 20 m/s
∴ Striking Time = 2 seconds