Question

For a ground-to-ground projectile, an object is at point A at t = $\frac{T}{3}$ , at point B at t = $\frac{5 T}{6}$ and reaches the
ground at t = T. The difference in heights between points A and B is

Moderate

Solution

As, $T = \frac{2 u_{y}}{g} \Rightarrow u_{y} = \frac{g T}{2}$
$\Rightarrow h_{A} = u_{y} t_{A} - \frac{1}{2} g t_{A}^{2} = u_{y} (\frac{2 u_{y}}{3 g}) - \frac{1}{2} g {(\frac{2 u_{y}}{3 g})}^{2}$
$= \frac{4}{9} \frac{u_{y}^{2}}{g} = (\frac{4}{9 g}) {(\frac{g T}{2})}^{2} = \frac{g T^{2}}{9} (∵ u_{y} = \frac{g T}{2})$
$\Rightarrow h_{B} = u_{y} (\frac{5}{6} \times \frac{2 u_{y}}{9}) - \frac{1}{2} \times g \times {(\frac{5}{6} \times \frac{2 u_{y}}{g})}^{2}$
$= \frac{5}{18} \frac{u_{y}^{2}}{g} = \frac{5}{18 g} {(\frac{g T}{2})}^{2} = \frac{5}{72} g T^{2}$
$∴ h_{A} - h_{B} = \frac{g T^{2}}{24}$

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