Question

Q.33. A projectile is projected with certain velocity at an angle $\theta$ with the horizontal from the ground. If its time of flight is T then the magnitude of change in its velocity when body reaches to ground again is (g is acceleration due to gravity)
(1) gT
(2) $\frac{gT}{2}$
(3) 2gT
(4) $\frac{g^2}{T}$

Answer 1

Dear Student,
$$\begin{gathered} whenbodytouchesthegrounditwillmakeangle\theta with \\ horizontal \\ soanglebetweeninitialvelocityandfinalvelocityis2\theta \\ hencedifferenceintheirvelocities \\ ∆v=\sqrt{u^2+u^2-2u^2\cos \left(2\theta \right)}=2u\sin \left(\theta \right) \\ butaccordingtoquestion \\ timeofflight=T=\frac{2u\sin \left(\theta \right)}{g} \\ \Rightarrow 2u\sin \left(\theta \right)=gT \\ hence \\ ∆v=gT \end{gathered}$$

Regard