Figure shows two positions $A$ and $B$ at the same height $h$ above the ground. If the maximum height of the projectile is $H$ , then determine the time $t$ elapsed between the positions $A$ and $B$ in terms of $H$ .

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Solution

Let $T$ be the time of flight. We can now write
$T = \frac{2 u sin θ}{g}$
$T^{2} = \frac{4 u^{2} s i n^{2} θ}{g^{2}} = \frac{8}{g} (\frac{u^{2} s i n^{2} θ}{g}) = \frac{8 H}{g}$
Hence, we can write in similar way, $t^{2} = \frac{8 (H - h)}{g}$
Thus, $t = \sqrt{\frac{8}{g}} (H - h)$

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Figure shows two positions A and B at the same height h above the ground. If the maximum height of the projectile is H, then determine the time t elapsed between the positions A and B in terms of H.

Let T be the time of flight. We can now writeT=2usinθgT2=4u2sin2θg2=8g(u2sin2θg)=8HgHence, we can write in similar way, t2=8(H−h)gThus, t=√8g(H−h)

Figure shows two positions $A$ and $B$ at the same height $h$ above the ground. If the maximum height of the projectile is $H$ , then determine the time $t$ elapsed between the positions $A$ and $B$ in terms of $H$ .

Let $T$ be the time of flight. We can now write
$T = \frac{2 u sin θ}{g}$
$T^{2} = \frac{4 u^{2} s i n^{2} θ}{g^{2}} = \frac{8}{g} (\frac{u^{2} s i n^{2} θ}{g}) = \frac{8 H}{g}$
Hence, we can write in similar way, $t^{2} = \frac{8 (H - h)}{g}$
Thus, $t = \sqrt{\frac{8}{g}} (H - h)$