Question

Two bodies are thrown vertically upward, with the same initial velocity of $98m/s$ but $4sec$ apart. How long after the first one is thrown when they meet ?

• A. $10sec$
• B. $11sec$
• C. $12sec$
• D. $13sec$

Answer 1

The correct option is C $12sec$

Both the bodies meet each other at t secs after the first one is thrown

Displacement covered by first object in t secs $s_1=ut-\frac{1}{2}g{t}^2=98t-\frac{1}{2}9.8\times t^2$

Displacement covered by second object in (t-4) secs $s_2=98(t-4)-\frac{1}{2}9.8\times (t-4{)}^2$

Since they meet $s_1=s_2$

We get $t=12secs$