Question

A ball is dropped from a height of 20 m a second ball is thrown downwards from the same height after 1 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

For First ball:

Height = 20 m

Acceleration = 9.8  m/s^2

Initial velocity (u) = 0 m/s

Time taken = t

$H = ut + \frac{1}{2}\times a \times t^2$

$20 = 0 + \frac{1}{2} \times 9.8 \times t^2$

t = 2.02 seconds

For second ball:

Initial velocity = u

Time taken = 2.02 - 1 = 1.02 seconds

a = 9.8 m/s^2

Height (h) = 20 m

$H = ut + \frac{1}{2}\times a \times t^2$

$20 = u \times 1.02 + \frac{1}{2}\times 9.8 \times 1.02^2$

$u = \frac{14.9}{1.02}$

u = 14.6 m/s

Hence, the initial velocity of second ball is 14.6 m/s

Answer 2

Answer:

For First ball:

Height = 20 m

Acceleration = 9.8  m/s^2

Initial velocity (u) = 0 m/s

Time taken = t

t = 2.02 seconds

For second ball:

Initial velocity = u

Time taken = 2.02 - 1 = 1.02 seconds

a = 9.8 m/s^2

Height (h) = 20 m

u = 14.6 m/s

Hence, the initial velocity of second ball is 14.6 m/s

Explanation: