Question

A man throws a ball vertically upward and it rises through $40 m$ and returns to his hands, what was the ascent initial velocity of the ball and for how much time $(T)$ it remained in the air?

• A. $T=5$ sec, $u=20 \sqrt{3}$ m/s
• B. $T=5.64$ sec, $u=10 \sqrt{2} m / s$
• C. $T=5.64$ sec, $u=20 \sqrt{2} m / s$
• D. None of these

Answer 1

Answer: (C)

Velocity of descent or ascent always equal.
$V=\sqrt{2 \times g \times h}=\sqrt{2 \times 10 \times 40}=\sqrt{800}=20 \sqrt{2} m / s$
Time of ascent and descent always equal.
$T =$ time of ascent $+$ time of descent
$T = t _{1}+ t _{2}$
Let $t$, is the time of ascent and $t,$ be that of descent
$v = u - g t$
$0= u - gt _{1} \quad v = u + gt _{2}$
$u = g t _{1} \quad u = 0 + gt _{2}$
$t_{1}=\frac{u}{g} \quad \frac{u}{g}=t_{2}$
$T = t _{1}+ t _{2}$
$T=\frac{u}{g}+\frac{u}{g} $
$=\frac{2 u}{g}=\frac{2 \times \sqrt{800}}{10}$
$T=4 \times \sqrt{2}$ sec
$T=4 \times 1.41=5.64$ seconds