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A particle is projected vertically upwards from a point $A$ on the ground. It takes time $t_1$ to reach a point $B$, but it still continues to move up. If it takes further $t_2$ time to reach the ground from point $B$. Then height of point $B$ from the ground is-
$\frac{1}{2} g(t_1 + t_2)^2$
$g\, t_1\, t_2$
$\frac{1}{8} g(t_1 + t_2)^2$
$\frac{1}{2} g t_1 \, t_2$
Solution
$\mathrm{h}=\mathrm{ut}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2}$
$\mathrm{O}=\mathrm{u}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)-\frac{1}{2} \mathrm{g}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}$
so $\mathrm{u}=\mathrm{g}\left(\frac{\mathrm{t}_{1}+\mathrm{t}_{2}}{2}\right)$
$\mathrm{h}=\mathrm{g}\left(\frac{\mathrm{t}_{1}+\mathrm{t}_{2}}{2}\right) \mathrm{t}_{1}-\frac{1}{2} \mathrm{gt}_{1}^{2}$
$=\frac{\mathrm{gt}_{1} \mathrm{t}_{2}}{2}$
