First slide

Projection Under uniform Acceleration

Question

A shell of mass 2 m fired with speed u at an angle $θ$ to the horizontal explodes at the highest point of its motion into two fragments of mass m each. If one fragment, whose initial speed is zero, falls vertically, .the distance at which the other fragment falls from the gun is given by

Easy

A

$\frac{3}{2} \frac{u^{2} \sin 2 θ}{g}$
B

$\frac{u^{2} \sin 2 θ}{g}$
C

$\frac{2}{3} \frac{u^{2} \sin 2 θ}{g}$
D

$\frac{3}{2} \frac{u^{2} \sin^{2} θ}{g}$

Solution

Momentum of the shell at highest point
$= 2 m (ucos θ)$
As one fragment has initial speed zero (stationary) and hence the other fragment of mass m has velocity component $(2 ucos θ)$ in horizontal direction. This will move under gravity. So it wilt fall at a distance which is equal to range.
$R = \frac{u^{2} \sin 2 θ}{g}$
Hence, distance covered from the gun
$\begin{array}{l} = \frac{R}{2} + R = \frac{3}{2} R \\ = \frac{3}{2} [\frac{u^{2} \sin 2 θ}{g}] \end{array}$

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