Question

A child drops a ball from a height of 10m. Assume that its velocity increases uniformly at the rate of 10m/s2 . Find
(i) the velocity with which the ball strikes the ground
(ii) the time taken by the ball to reach the ground

Answer 1

The distance here is 10 m per s

And the acceleration is 10 m per s2 v²=u²+2as

v²= 0²+ 2×10×10 v²= 200 v= 10√2 v=u+at 10√2=0+10t => 10√2-10=t => 10 (√2-1)= t => 4.1

Here the time is 4.1 seconds And the velocity is 10 sq root of 2

Answer 2

   ANSWER    

Acceleration(a) = 10 m/s²

Initial Velocity(u) = 0 m/s

Final velocity be 'v' m/s

g = 10 m/s²

Now Using equation of motion:

v² + u² = 2gh

v = √2gh

  = √200

  =  10√2 m s⁻¹

Now, Let the time be 't' sec.

.

Equation of motion:

h = ut + 1/2gt²

h = 1/2gt²

t = √(2h/g)

t = √2 sec

∴ Velocity = 10√2 m s⁻¹

∴ Time = √2 second