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

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(D) The vertical displacement of a projectile is given by $h(t) = u_y t - \frac{1}{2} g t^2$,where $u_y$ is the initial vertical velocity.Given the to

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

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(D) The vertical displacement of a projectile is given by $h(t) = u_y t - \frac{1}{2} g t^2$,where $u_y$ is the initial vertical velocity.Given the total time of flight is $T$,we have $T = \frac{2 u_y}{g}$,which implies $u_y = \frac{g T}{2}$.At $t_A = \frac{T}{3}$,the height $h_A$ is:$h_A = \left(\frac{g T}{2}\right) \left(\frac{T}{3}\right) - \frac{1}{2} g \left(\frac{T}{3}\right)^2 = \frac{g T^2}{6} - \frac{g T^2}{18} = \frac{3 g T^2 - g T^2}{18} = \frac{2 g T^2}{18} = \frac{g T^2}{9}$.At $t_B = \frac{5T}{6}$,the height $h_B$ is:$h_B = \left(\frac{g T}{2}\right) \left(\frac{5T}{6}\right) - \frac{1}{2} g \left(\frac{5T}{6}\right)^2 = \frac{5 g T^2}{12} - \frac{25 g T^2}{72} = \frac{30 g T^2 - 25 g T^2}{72} = \frac{5 g T^2}{72}$.The difference in heights is $H = h_A - h_B = \frac{g T^2}{9} - \frac{5 g T^2}{72} = \frac{8 g T^2 - 5 g T^2}{72} = \frac{3 g T^2}{72} = \frac{g T^2}{24}$.

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

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