Question

A ball is projected vertically upward with an initial velocity of $50 \,ms ^{-1}$ at $t =0 \,s$. At $t =2 \,s$. another ball is projected vertically upward with same velocity. At $t =$ ______$s$, second ball will meet the first ball $\left( g =10 \,ms ^{-2}\right)$.

Answer 1

Let they meet at $t = t$
So first ball gets $t \sec$.
& $2^{\text {nd }}$ gets $( t -2) \sec$ . & they will meet at same height
$h _{1}=50 t -\frac{1}{2} gt ^{2}$
$h _{2}=50( t -2)-\frac{1}{2} g ( t -2)^{2}$
$h _{1}= h _{2}$
$50 t -\frac{1}{2} gt ^{2}=50( t -2)-\frac{1}{2} g ( t -2)^{2}$
$100=\frac{1}{2} g \left[ t ^{2}-( t -2)^{2}\right]$
$100=\frac{10}{2}[4 t -4]$
$5= t -1$
$t =6 \sec .$