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Q.
A stone thrown at an angle $ \theta $ to the horizontal reaches a maximum height $H$ . Then the time of flight of stone will be
Punjab PMETPunjab PMET 2005Motion in a Plane
Solution:
Maximum height
$H=\frac{u^{2} \sin ^{2} \theta}{2 g}$ ...(1)
Time of flight
$T =\frac{2 u \sin \theta}{g}$
$\Rightarrow u \sin \theta =\frac{T g}{2}$ ...(2)
From equations (1) and (2), we get
$H =\frac{1}{2 g}\left(\frac{T g}{2}\right)^{2}$
$H=\frac{T^{2} g^{2}}{8 g}$
$T^{2} =\frac{8 H}{g}$
$T=2 \sqrt{\left(\frac{2 H}{g}\right)}$