First slide

Projectile motion

Question

Two seconds after projection a projectile is travelling in a direction inclined at 30^o to the horizontal after one more second, it is travelling horizontally, the magnitude and direction of its velocity are

Moderate

A

$2 \sqrt{20} m / s, 60^{\circ}$
B

$20 \sqrt{3} m / s, 60^{\circ}$
C

$6 \sqrt{40} m / s, 30^{\circ}$
D

$40 \sqrt{6} m / s, 30^{\circ}$

Solution

Let in 2 seconds, the body reaches up to point A and after one more second up to point B.
Total time of ascent for a body is given 3sec, $i.e., t = \frac{u \sin θ}{g} = 3$
$∴ u \sin θ = 10 \times 3 = 30 . . . . . . . (i)$
Horizontal component of velocity remains always constant
$u \cos θ = v \cos 30^{\circ} . . . . . . . . . (i i)$
For vertical upward motion between point O and A,
$\begin{array}{l} v \sin 30^{\circ} = u \sin θ - g \times 2 [Using v = u - g t] \\ v \sin 30^{\circ} = 30 - 20 [As u \sin θ = 30] \\ ∴ v = 20 m / s \end{array}$
Substituting this value in equation (ii),
$u \cos θ = 20 \cos 30^{\circ} = 10 \sqrt{3} . . . . . . . . . . (i i i)$
$From equations (i) and (iii), u = 20 \sqrt{3} and θ = 60^{\circ}$

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