Q: A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 s, another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v ? ( Take g = 10 ms–2)

(a) 75 m s^{–1}

(b) 55 ms^{–1}

(c) 40 ms^{–1}

(d) 60 ms^{–1}

Ans: (a)

As two balls will meet , hence

Distance covered by first ball in 18 sec = Distance covered by second ball in 12 sec

$ \displaystyle \frac{1}{2}g (18)^2 = v \times 12 + \frac{1}{2}g(12)^2$

$\displaystyle \frac{1}{2}g (18)^2 – \frac{1}{2}g(12)^2 = v \times 12 $

$ \displaystyle \frac{1}{2}\times 10 \times (18+12)(18-12) = v \times 12 $

$ \displaystyle 5 \times (30)(6) = v \times 12 $

$ \displaystyle v = \frac{900}{12} $

v = 75 m/s