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 gT26x. Find value of x.

T=2uyg∴uy=gT2 hA=uytA−12gtA2=uy2uy3g−12g2uy3g2=49uy2g=49ggT22=gT29hB=uy56×2uyg−12×g×56×2uyg2=518uy2g=518ggT22=572gT2∴ hA−hB=gT224.Hence x = 24/6 = 4.

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 gT26x. Find value of x.

T=2uyg∴uy=gT2 hA=uytA−12gtA2=uy2uy3g−12g2uy3g2=49uy2g=49ggT22=gT29hB=uy56×2uyg−12×g×56×2uyg2=518uy2g=518ggT22=572gT2∴ hA−hB=gT224.Hence x = 24/6 = 4.

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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 $\frac{{gT}^{2}}{6 x}$ . Find value of x.

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Detailed Solution

$\begin{array}{l} T = \frac{2 u_{y}}{g} \\ ∴ u_{y} = \frac{gT}{2} \\ h_{A} = u_{y} t_{A} - \frac{1}{2} {gt}_{A}^{2} \end{array}$

$\begin{matrix} = u_{y} (\frac{2 u_{y}}{3 g}) - \frac{1}{2} g {(\frac{2 u_{y}}{3 g})}^{2} \\ = \frac{4}{9} \frac{{u_{y}}^{2}}{g} = (\frac{4}{9 g}) {(\frac{gT}{2})}^{2} = \frac{{gT}^{2}}{9} \\ h_{B} = u_{y} (\frac{5}{6} \times \frac{2 u_{y}}{g}) - \frac{1}{2} \times g \times {(\frac{5}{6} \times \frac{2 u_{y}}{g})}^{2} \\ = \frac{5}{18} \frac{u_{y}^{2}}{g} = \frac{5}{18 g} {(\frac{gT}{2})}^{2} = \frac{5}{72} {gT}^{2} \\ ∴ h_{A} - h_{B} = \frac{{gT}^{2}}{24} . \end{matrix}$