A ground to ground projectile is at point A at t=T3, is at point B at t=5T6 and reaches the ground at t=T. The difference in heights between points A and B is

T=2uyg ⇒ uy=gT2 hA=uytA-12gtA2 =uy2uy3g-12g2uy3g2 =49uy2g=49ggT22=gT29 Similarly, hB=uy56×2uyg-12×g×56×2uyg2 =518uy2g=518ggT22=572gT2 ∴ hA-hB=124gT2

A ground to ground projectile is at point A at t=T3, is at point B at t=5T6 and reaches the ground at t=T. The difference in heights between points A and B is

T=2uyg ⇒ uy=gT2 hA=uytA-12gtA2 =uy2uy3g-12g2uy3g2 =49uy2g=49ggT22=gT29 Similarly, hB=uy56×2uyg-12×g×56×2uyg2 =518uy2g=518ggT22=572gT2 ∴ hA-hB=124gT2

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A ground to ground projectile is at point A at $t = \frac{T}{3}$ , is 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

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$\frac{g T^{2}}{6}$

$\frac{g T^{2}}{18}$

none of these

$\frac{g T^{2}}{12}$

answer is D.

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Detailed Solution

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