Question

A ball is dropped from the top of a tower 30 height at the same instant second ball is thrown upwards from the ground with an initial velocity of 15 metre per second when and where do they cross each other and with what relative velocity

Answer 1

$\textbf{ \huge{ Hello Mate! }}$

Let's think of a ball dropped from up.

It would have u = 0 ( Since ball had no velocity before ).

So the distance of ball travelled from A to c would be x.

$s = ut + \frac{1}{2} a {t}^{2} \\ x = (0)(t) + \frac{1}{2} a {t}^{2} \\ x = \frac{1}{2} a {t}^{2} ...(1)$

Now, lets think for a ball which was thrown from down. It u = 15 m/s ( Given ). So if whole tower is 30 m high and AC is x then BC = 30 - x.

$30 - x = (15)(t) + \frac{1}{2} ( - a){t}^{2} \\ 30 - x = 15t - \frac{1}{2} a{t}^{2} ....(2)$

Doing eq (2) + (1) we get

30 = 15t

2 s = time.

x = 1 / 2 × 10 × 2 × 2

x = 20 m ( from top )

Case for ball went down.

v = u + at
v = 0 + (10)(2)
v = 20 m/s

Relative velocity = 20 m/s + 15 m/s = 35 m/s

$\textsf{\red{ (a) 2s ; (b) 20 m from top : R Velocity = 35 m/s }}$

$\textbf{ Have great future ahead! }$