A ground to ground projectile is at point A a | vedclass

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Question

(d)

$T=\frac{2 u_y}{g}$

$u_y =\frac{g T}{2}$

$h_A =u_y t_A-\frac{1}{2} g t_A^2$

$=u_y\left(\frac{2 u_y}{3 g}\right)-\frac{1}{2} g\left(\fr

Vedclass · Accepted Answer

$\frac{g T^2}{24}$

Vedclass · Answer

(d)

$T=\frac{2 u_y}{g}$

$u_y =\frac{g T}{2}$

$h_A =u_y t_A-\frac{1}{2} g t_A^2$

$=u_y\left(\frac{2 u_y}{3 g}ight)-\frac{1}{2} g\left(\frac{2 u_y}{3 g}ight)^2$

$=\frac{4}{9} \frac{u^2 y}{g}=\left(\frac{4}{9 g}ight)\left(\frac{g T}{2}ight)^2=\frac{g T^2}{9}$

$h_{ B } =u_y\left(\frac{5}{9} 	imes \frac{2 u_y}{g}ight)-\frac{1}{2} 	imes g 	imes\left(\frac{5}{6} 	imes \frac{2 u_y}{g}ight)^2$

$	herefore =\frac{5}{18} \frac{u_y^2}{g}=\frac{5}{18 g}\left(\frac{g T}{2}ight)^2=\frac{5}{72} g T^2$

$	herefore H=h_A-h_B$

$\Rightarrow \frac{g T^2}{9}-\frac{5 g T^2}{72}$

$\Rightarrow \frac{3 g T^2}{72}$

$	herefore H=\frac{g T^2}{24}$

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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