Question

A particle is given horizontal projection from the top of a tower with the speed 20 m/s. What will be the angle of its velocity with horizontal after 2s (g = 10m/$s^2$)

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. $ta{n}^{-1}\frac{5}{4}$

Answer 1

The correct option is B ${45}^{\circ }$

The correcct option is B.

A mparticle is projected horizontal and the initial horizontal component of the projected particle is $V_h=20m/s$

The initial vertical component of the projected particle is $V_v=0$,

This will change with time during its flight by the action of the gravitional force will be:

$V_v^{\prime }=V_v+t\times g$

$=0+10\times 2=20m/s$

Since $g=10m/s^2$

so,

$\frac{V_v^{\prime }}{V_h}=tan\theta$

$\theta =ta{n}^{-1}\frac{20}{20}$

$\theta =ta{n}^{-1}1={45}^0$