Question

From the top of a tower of 100m height, a ball is dropped and at the same time another ball is projected vertically upwards from the ground with a velocity of 25ms^-1. Find when and where the two balls will meet. Take g = 9.8 ms^-2.

Answer 1

Let (A) be the top of a tower and (B) beits foot. Let two balls meet at (C) after time (t).

Let AC = X, then BC = 100 − x.

Taking vertical downward motion of the ball dropped from the top, we have

u = 0, a = 9.8 ms⁻², S = xt = t

As,

\(S = ut + \frac 12 at^2\)

\(\therefore x = 0 + \frac 12 \times 9.8 \times t^2\) 

\(= 4.9t^2\)    .......(i)

Taking vertical upward motion of the ball thrown up from (B), we have

u = 25ms⁻¹, a = −9.8ms⁻²,

S = (100 − x), t = t

As,

\(S = ut + \frac 12 at^2\)

\(\therefore 100 - x = 25 t + \frac 12 (-9.8)t^2\)

\(= 25t - 4.9t^2\)    ......(ii)

Adding (i) and (ii) we have ,

100 = 25t or t = 4s

Putting this value in (i) we get,

x = 4.9 × 16 = 78.4 m

Hence, the two balls will meet after 4 seconds at a distance 78.4m below the top.

Answer 2

Let AC = x

Then BC = 100 - x

x = ut + \(\frac {1}{2}\) g t²

x = 4.9 t² ......... (1)

BC = ut - \(\frac {1}{2}\) g t²

= 25 t - \(\frac{1}{2}\) × 9.8 t²

100 - x = 25 t - 4.9 t² ........ (2)

Substituting 4.9 t² = x from (1) in (2)

100 - 2 = 25 t - x

⇒ t = 4 seconds

x = 4.9 t²

= 4.9 × 4²

= 78.4 m from the top or 21.6 m from the ground.