Question

a ball is gently dropped from a height of 20m if its velocity increases uniformly at the rate of 10m/s with what velocity will it strike the ground .​

Answer 1

Answer:

Answer is 20m/s

Explanation:

Here,   Height from which ball is dropped (s) = 20 m

Now,    Rate of change of velocity w.r.t. time is give as 10 m/s

   ⇒ Acceleration of body (a) = 10 m/s

Also,  The ball is gently dropped.

  ⇒ Initial velocity (u) = 0

Now, Using $3^{rd}$ equation of motion,

  $v^{2} - u^{2} = 2as$         ⇒     $v^{2} - 0 = 2*10*20$

 ⇒   $v^{2} = 400$    ⇒      $v = 20 m/s$

Answer 2

Answer:

20 ms^-1

Explanation:

Given

Distance = 20m

Acceleration = 10ms^-2

initial velocity = 0

Now,

2as = v^2 - u^2

2×10ms^-2×20m = v^2 - 0^2

400ms^-2 = v^2

√400s^-2 = v

20ms-1 = v

Hope it helps !!!!