Question

A rubber ball is shot straight up from the ground with a speed of 12 m/s. Simultaneously, a second identical rubber ball is dropped from rest exactly 7 m directly above the first ball. At what height (in m) do the two balls collide?

Answer 1

Answer:

6.17 m

Explanation:

We are given that

Initial speed of rubber ball, u=12 m/s

Total height, h=7 m

Initial speed of second ball, u'=0

We have to find the height at which the two balls collide.

Let  first rubber ball and second ball strikes after t time.

For first ball

Distance traveled by first ball in time t

$S=ut-\frac{1}{2}gt^2$

Substitute the value

$S=12t-\frac{1}{2}(9.8)t^2$

$S=12t-4.9t^2$   ...(1)

Distance traveled by second ball in time t

$7-S=\frac{1}{2}(9.8)t^2$

$7-S=4.9t^2$  .....(2)

Using equation (2) in equation (1) we get

$S=12t-(7-S)$

$S=12t-7+S$

$\implies 12t=7$

$t=\frac{12}{7}$sec

Now, using the value of t

$S=12(\frac{12}{7})-4.9(\frac{12}{7})^2$

S=6.17 m

Hence, at height 6.17 m the two balls collide .