Question

A ball is dropped freely while another is thrown vertically downward with an initial velocity $\frac{v}{2}$ from the same point simultaneously. After 't' second they are separated by a distance of?

• A. $\frac{vt}{2}$
• B. $\frac{1}{2}g{t}^2$
• C. $vt$
• D. $vt+\frac{1}{2}g{t}^2$

Answer 1

The correct option is A $\frac{vt}{2}$

From Newton equation of motion we know

$s=ut+\frac{1}{2}g{t}^2$

For ball 1 distance traveled in t seconds will be

$s_1=\frac{1}{2}g{t}^2$ as initial velocity = 0 m/s

and for ball 2 distance traveled in t seconds will be

$s_2=ut+\frac{1}{2}g{t}^2$ in this case initial velocity u= $\frac{v}{2}$ m/s

$=\frac{vt}{2}+\frac{g{t}^2}{2}$

Now subtracting $s_1-s_2$ we get

$s_2-s_1=\frac{vt}{2}+\frac{g{t}^2}{2}-\frac{g{t}^2}{2}{s}_2-s_1=\frac{vt}{2}$