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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
$\frac{g T^2}{6}$
$\frac{g T^2}{12}$
$\frac{g T^2}{18}$
$\frac{g T^2}{24}$
Solution
(d)
$T=\frac{2 u_y}{g}$
$u_y =\frac{g T}{2}$
$h_A =u_y t_A-\frac{1}{2} g t_A^2$
$=u_y\left(\frac{2 u_y}{3 g}\right)-\frac{1}{2} g\left(\frac{2 u_y}{3 g}\right)^2$
$=\frac{4}{9} \frac{u^2 y}{g}=\left(\frac{4}{9 g}\right)\left(\frac{g T}{2}\right)^2=\frac{g T^2}{9}$
$h_{ B } =u_y\left(\frac{5}{9} \times \frac{2 u_y}{g}\right)-\frac{1}{2} \times g \times\left(\frac{5}{6} \times \frac{2 u_y}{g}\right)^2$
$\therefore =\frac{5}{18} \frac{u_y^2}{g}=\frac{5}{18 g}\left(\frac{g T}{2}\right)^2=\frac{5}{72} g T^2$
$\therefore H=h_A-h_B$
$\Rightarrow \frac{g T^2}{9}-\frac{5 g T^2}{72}$
$\Rightarrow \frac{3 g T^2}{72}$
$\therefore H=\frac{g T^2}{24}$
