Question

A ball is dropped from the top of a building and simultaneously, another ball is projected vertically upward from the ground with a velocity of $25m/s$. If the initial distance between these balls is $100m$. Calculate where the two balls will meet above the ground.
(Take g = $10m/s^2$ )

• A. $20m$
• B. $40m$
• C. $60m$
• D. $80m$

Answer 1

The correct option is A $20m$

We consider that the ball dropped from the building is $b_1$ and another ball, projected vertically upward is $b_2$.
Both the balls have an acceleration of g in the downward direction and they meet after ‘t’ sec.

Initial velocity of ball($b_1$) = $0m/s$

Initial velocity of ball($b_2$) = $25m/s$

Distance covered by ball($b_2$) in time t = $x$

Total height of the cliff(H) = $100m$

●Distance covered by ball($b_1$) in time t = $100-x$

Plug in values in second equation of motion

For Ball ($b_1$) :

$100-x$ = $0$ + $\frac{1}{2}$$g{t}^2$

$100-x$ = $\frac{1}{2}$$g{t}^2$ ....(1)

For Ball(b2) :

$x$ = $u_2$$t$ - $\frac{1}{2}$$g{t}^2$ …(2)

On adding (1) & (2), we get the final equation :

100 =$\frac{1}{2}$$g{t}^2$ +$u_2$$t$ - $\frac{1}{2}$$g{t}^2$

100 = $u_2$$t$

t = 4 s

Substituting t = 4 in equation 1

$x$ = (25)(4) - $\frac{1}{2}$($10\ast 4^2$)

$x$ = $20m$