First slide

Vertical projection from ground

Question

A particle thrown up vertically reaches its highest point in time t₁ and returns to the ground in time t₂.
The air resistance exerts a constant force on the particle opposite to its direction of motion, then

Moderate

A

$t_{1} > t_{2}$
B

$t_{1} = t_{2}$
C

$t_{1} < t_{2}$
D

none of above

Solution

See figure

For upward motion
$0 = u - a_{1} t_{1} or u = a_{1} t_{1}$
$\begin{array}{ll} and h = {ut}_{1} - \frac{1}{2} a_{1} {t_{1}}^{2} \\ ∴ h = (a_{1} t_{1}) t_{1} - \frac{1}{2} a_{1} {t_{1}}^{2} = \frac{1}{2} a_{1} {t_{1}}^{2} \end{array}$
For downward motion
$\begin{array}{l} h = \frac{1}{2} a_{2} t_{2} \\ ∴ \frac{t_{2}}{t_{1}} = \sqrt{(\frac{a_{1}}{a_{2}})} = \sqrt{\frac{[g + (R / m)]}{[g - (R / m)]}} \\ or t_{1} < t_{2} \end{array}$

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