Motion In A Straight Line Ques 34
34. A ball is dropped from a high rise platform at $t=0$ starting from rest. After 6 seconds 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 $v$ ?
[2010] (take $g=10 m / s^{2}$ )
(a) $75 m / s$
(b) $55 m / s$
(c) $40 m / s$
(d) $60 m / s$
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Answer:
Correct Answer: 34.(a)
Solution:
- (a) Clearly distance moved by $1^{\text{st }}$ ball in $18 s=$ distance moved by $2^{\text{nd }}$ ball in $12 s$…(i)
Now, distance moved in $18 s$ by $1^{\text{st }}$
ball $=\frac{1}{2} \times 10 \times 18^{2}=90 \times 18=1620 m$
Distance moved in $12 s$ by $2^{\text{nd }}$ ball
$=u t+\frac{1}{2} g t^{2} \Rightarrow 12 v+\frac{1}{2} \times 10 \times(12)^{2}$ $\Rightarrow \quad 12 v+720$
From equation (i)
$\therefore \quad 1620=12 v+5 \times 144$
$\Rightarrow v=135-60=75 ms^{-1}$
BETA
