Question

A ball is dropped gently from a height of 20m. If it's velocity increase uniformly at the rate of 10m^-2 with what velocity will it strike the ground? After what time will it strike the ground?

Answer 1

Answer:

final velocity = 20m/s, time taken = 2 sec

Explanation:

here,

        s (h) = 20m

        a = 10m/$s^{2}$

        u = 0

using equation, $v^{2} - u^{2} = 2as$

$v^{2} - 0^{2}$ = 2×10×20

⇒$v^{2}$ = 400

⇒v = 20

Now,

      using equation, v = u + at

20 = 0 + 10×t

⇒10t = 20

⇒t = 20/10

⇒t = 2

Answer 2

Answer:

Given, initial velocity of ball, u=0

Final velocity of ball, v=?

Distance through which the balls falls, s=20m

Acceleration a=10ms −2

Time of fall, t=?

We know

v 2 −u 2

=2as

or v 2 −0=2×10×20=400 or v=20ms −1

Now using v=u+at we have

20=0+10×t or t=2s