Question

17. A ball is dropped from a height of 20 m. A second ball is thrown down wards from
the same height after one second with initial velocity u. If both the balls reach the
ground at the same time, calculate the initial velocity of the second ball.​

Answer 1

Answer:

For first ball,

Given height h =20m

Initial velocity u =0

Acceleration due to gravity = 10 m/s

Time taken by ball to fall height h = t

Using,

h = ut +0.5 gt

Here plug in h, u, g to find t =.........?

20 = 0.5 *10 *t

=> 20 = 5t

=> t^2 =4

=> t = 2 s

for 2nd ball

u' =.....?

g=10 m/s

Time taken by 2nd ball to fall a height h t' = (t-1) = 2-1 = 1s

h = 20m

Again,

h = u't' +0.5g(t')^2

here plug in t',h, g to find u' =........?

20 = u'(1) +0.5 (10) *(1)^2

=> 20 = u' + 5

u' = 20-5 = 15 m/s

Answer 2

Answer:

1. The initial velocity of the second ball = 20m/s

b.) The balls hit the surface with same velocity = 2s

EXPLANATION ::

By the Given condition

Initial velocity of ball = u=0

Final velocity of ball= v (say)

Distance =20m

Acceleration a=10m/s^2

Let Time of fall = t

We know that

v ^2−u^2=2as

v^2−0=2×10×20=400

v=20m/s

Now v=u+at gives

20=0+10×t

t=2s