Question

Two balls are dropped from different heights at different instants. Second ball is dropped $2$sec after the first ball. If both balls reach the ground simultaneously after $5$sec. of dropping the first ball, the difference of initial heights of the two balls will be: $(g=9.8m/s^2)$.

• A. $58.8$m
• B. $78.4$m
• C. $98.0$m
• D. $117.6$m

Answer 1

The correct option is B $78.4$m

For first ball $S=ut+\frac{1}{2}g{t}^2\therefore (t=5sec)$

$=0+\frac{1}{2}g{\left(5\right)}^2=\frac{25g}{2}=H_1$

For second ball

$S=ut+\frac{1}{2}g{t}^2\therefore (t=5-2=3sec)$

$=0+\frac{1}{2}g{\left(3\right)}^2=\frac{9g}{2}=H_2$

Difference between the initial heights

$H_1-H_2=\frac{25g}{2}-\frac{9g}{2}$

$=\frac{16g}{2}$

$=8g=8\times 9.8$

$=78.4$

$\therefore$ The difference of initial heights of the two balls is $78.4m$