Question

A ball thrown up vertically returns to the thrower after $6s$. Find the position of the ball after $4s$.
(Take acceleration due to gravity as $9.8m{s}^{-2}$)

• A. $39.2m$
• B. $41.2m$
• C. $44.4m$
• D. $29.2m$

Answer 1

The correct option is A $39.2m$

Given:
Acceleration due to gravity, $g=-9.8m{s}^{-2}$
Total time taken by the ball, $t=6s$
(Upward direction is taken as positive)

Let the initial velocity of the ball be $u$.
After complete travel, final position is same as the initial. Hence, displacement, $s=0$

From equation of motion:
$s=ut-\frac{1}{2}g{t}^2$
$u=\frac{1}{t}(s+\frac{1}{2}g{t}^2)$

Again, using equation of motion for the initial point and time, $t_2=4s$.
$s_2=u{t}_2-\frac{1}{2}g{t}_2^2$

Putting equation of $u$ in above equation, we get:
$s_2=\frac{t_2}{t}(s+\frac{1}{2}g{t}^2)-\frac{1}{2}g{t}_2^2$
Substituting values, we get:
$s_2=\frac{4}{6}(0+\frac{1}{2}\times 9.8\times 6^2)$$-\frac{1}{2}\times 9.8\times 4^2$
$s_2=39.2m$