First slide

Projection Under uniform Acceleration

Question

A particle is projected at an angle of $45^{o}$ . After 1 sec, it breaks into two equal parts. One part stops and the other part attains the height of 20 m after the breaking of the particle. Find the velocity of projection ( g = 10 $m / s^{2}$ )

Moderate

A

20 m/s
B

$20 \sqrt{2} m / s$
C

$10 \sqrt{2} m / s$
D

15 m/s

Solution

$\frac{V_{y}^{2}}{2 g} = 20 m$
$\Rightarrow V_{y} = 20 m / s$ = vertical velocity of the second part.
From conservation of momentum $m \times 0 + {mV}_{y} = (2 m) V_{y_{o}}$
$V_{y_{0}}$ vertical velocity of particle after one second before breaking of it
= $\frac{V_{y}}{2} = \frac{20}{2} = 10 m / s$
But $V_{y_{0}} = 10 = u_{y} - g t$
$\Rightarrow u_{y} = 10 + 10 \times 1 = 20 m / s$
Now, $u_{y} = u \sin 45^{o}$
$\Rightarrow u = \frac{u_{y}}{\sin 45^{o}} = 20 \sqrt{2} m / s$

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