A ball is projected vertically upward with an initial velocity of… - Vidyadip

MCQ

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 $t= \; \dots \; s$, second ball will meet the first ball $\left( g =10 \; ms ^{-2}\right)$.

✓
$6$
B
$5$
C
$4$
D
$3$

✓

Answer

Correct option: A.

$6$

a
Let they meet at $t = t$

So first ball gets $t \; sec$.

and $2^{\text {nd }}$ gets $( t -2) \; sec$ . and they will meet at same height

$h _{1}=50 t -\frac{1}{2} gt ^{2}$

$h _{2}=50( t -2)-\frac{1}{2} g ( t -2)^{2}$

$h _{1}= h _{2}$

$50 t -\frac{1}{2} gt ^{2}=50( t -2)-\frac{1}{2} g ( t -2)^{2}$

$100=\frac{1}{2} g \left[ t ^{2}-( t -2)^{2}\right]$

$100=\frac{10}{2}[4 t -4]$

$5= t -1$

$t =6 \; sec$

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