First slide

Projectile motion

Question

Velocity of a particle at time t = 0 is 2 m/s. A constant acceleration of 2 m/s² act on the particle for 2 s at an angle 60^o with the initial velocity. The magnitude of velocity of particle at the end of 2 s will be

Difficult

A

$2 \sqrt{5} m / s$
B

$2 \sqrt{7} m / s$
C

$2 \sqrt{6} m / s$
D

$2 \sqrt{3} m / s$

Solution

Let us make the components of initial velocity in the direction and perpendicular to the direction of acceleration as shown in the figure.
In the direction of acceleration:
Using $v = u + a t$ , we get
$\begin{array}{l} v_{1} = u \cos 60^{\circ} + a t \\ = 2 \times \frac{1}{2} + 2 \times 2 = 5 m / s \end{array}$
Velocity perpendicular to direction of acceleration remains same, which is
$u \sin 60^{\circ} = \sqrt{3} m / s$
So net velocity at the end of 2 s:
$v = \sqrt{v_{1}^{2} + (\sqrt{3})^{2}} = \sqrt{5^{2} + 3} = \sqrt{28} = 2 \sqrt{7} m / s$

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