Question

A ball of mass $100$ gm is projected vertically upward from the ground with a velocity of $50m{s}^{-1}$ . At the same time another identical ball is dropped from a height of $100$ m to fall freely along the same path as that followed by the first ball. After some time the two balls collide, stick together and finally fall to the ground. The time taken by the combined mass to fall to the ground is approximately $(g=10m{s}^{-2})$

• A. 4.5 s
• B. 6.5 s
• C. 9 s
• D. 13 s

Answer 1

The correct option is A 4.5 s

${V_1}^2=(50{)}^2-2gx$
${V_2}^2=2g(100-x)$
$\Rightarrow {V_2}^2=2g\left(100\right)+{V_1}^2-(50{)}^2$
$\Rightarrow {V_2}^2-{V_1}^2=2000-2500$
$\Rightarrow (V_1+V_2)(V_1-V_2)=\sqrt{-500}$
$V_1=50-gt$
$V_2=gt$
$\Rightarrow V_1+V_2=50$
$\Rightarrow V_1-V_2=+10$
$V_1=30m/s(\uparrow )$
$V_2=230m/s(\downarrow )$
$\Rightarrow x=\frac{900-2500}{20}=80$ m
$80=-5t+\frac{1}{2}g{t}^2$
$t^2-t-16=0$
$t=\frac{+1\pm \sqrt{1+64}}{2}=\frac{1\pm \sqrt{65}}{2}$
$=\frac{\sqrt{65}+1}{2}\approx 4.5$ sec