A body $A$ is projected upwards with a velocity of $98\,m/s$. The second body $B$ is projected upwards with the same initial velocity but after $4 \,sec$. Both the bodies will meet after...........$sec$
A body $A$ is projected upwards with a velocity of $98\,m/s$. The second body $B$ is projected upwards with the same initial velocity but after $4 \,sec$. Both the bodies will meet after...........$sec$
$6$
$8$
$10$
$12$
A body $A$ is projected upwards with a velocity of $98\,m/s$. The second body $B$ is projected upwards with the same initial velocity but after $4 \,sec$. Both the bodies will meet after...........$sec$
$h=u t-\frac{1}{2} g t^{2}$
Body 1: $h=98 t-\frac{1}{2} \times 9.8 \times t^{2}$
Body 2: $h=98(t-4)-\frac{1}{2} \times 9.8 \times(t-4)^{2}$
Equating the above two expressions we get:
$98 t-\frac{1}{2} \times 9.8 \times t^{2}=98(t-4)-\frac{1}{2} \times 9.8 \times(t-4)^{2}$
$\Rightarrow 98 \times 4+\frac{9.8}{2}(16-8 t)=0$
$\Rightarrow 392=-4.9(16-8 t)$
$\Rightarrow \frac{392}{4.9}=(8 t-16) \Rightarrow 80=(8 t-16) \Rightarrow 96=8 t \Rightarrow t=12 s$
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