Question

A body is projected vertically up to $t=0$ with a velocity of $98m/s$. Another body is projected from the same point with same velocity after $4$ seconds. Both bodies will meet after

• A. 6 s
• B. 8 s
• C. 10 s
• D. 12 s

Answer 1

Answer: (D)

12 s

Let they meet after time t from projection of first at height h above the ground the
$=ut-\frac{1}{2}g{t}^2$
for $1^{st}h=98t-\frac{1}{2}\times 9.8\times t^2......\left(i\right)$
for $2^{nd}h=98(t-4)-\frac{1}{2}\times 9.8\times (t-4{)}^2.....(ii)$
equation (i) and(ii)
$0=98\times 4+\frac{9.8}{2}(-8t+16)$
$\Rightarrow 40-4t+8=0\Rightarrow 4t=48\Rightarrow t=12s$