Question

A ball of mass 100 g is projected vertically upward from the ground with a velocity of 49 m/s. At the same time another ball is dropped from a height of 98 m to fall freely along the same path as the first ball. After some time the two balls collide and stick together and finally fall together.Find the time of flight of masses.

• A. 3.2 s
• B. 12.6 s
• C. 7.5 s
• D. 6.5 s

Answer 1

The correct option is D 6.5 s

Suppose they meet after time $t_0$ then $sum$ of magnitudes of their displacement should be $H=98m$

or $h_1+h_2=(49t_0-\frac{9.8}{2}{t}_0^2)+\frac{9.8}{2}{t}_0^2=98$ so $t_0=2sec$

Now we know that after $2sec$ the velocity of first particle in $upward$ direction will be $v_1=49-9.8\times 2=29.4m/s$

and the velocity of second particle after these 2sec will be $v_2=9.8\times 2=19.6m/s$ in $downward$ direction.

so the net $momentum$ in $upward$ direction will be $p=m\times 29.4-m\times 19.6=9.8m$

so net $upward$ velocity of htis combined mass$\left(2m\right)$ will be $V=\frac{p}{2m}=4.9m/s$

and this height from ground will be $h_1=49\times 2-4.9\times 4=78.4m$

now using following equation $h=-Vt+4.9t^2$ putting $h=78.4m$ and $V=4.9m/s$

we get $t=4.5sec$

so time of flight will be $t+t_0=6.5sec$