Question

A ball is projected with a speed ‘$v$’ at an angle ‘$\alpha$’ with the horizontal. Its speed when its direction of motion makes an angle ‘$\beta$’ with the horizontal.
A. $v\cos \alpha$
B. $v\cos \alpha \sec \beta$
C. $v\cos \beta sec\alpha$
D. $v\cos \alpha \cos \beta$

Answer 1

Use the kinematic expression for the final velocity in terms of acceleration and initial velocity of the particle. First determine the horizontal velocity of the projectile in the projectile motion. Also determine the vertical component of the velocity of the projectile using the kinematic equation. Hence, determine the required velocity of the particle.

Formulae used:
The expression for kinematic equation for the final velocity $v$ of the object is
$v = u + at$ …… (1)
Here, $u$ is initial velocity of the object, $a$ is acceleration of the object and $t$ is time.
The acceleration $a$ of an object is given by
$a = \dfrac{v}{t}$ …… (2)
Here, $v$ is the velocity of the object and $t$ is the time.

Complete Step by Step Answer:
We have given that the velocity of projection of the ball is $v$ and the angle of projection of the ball with the horizontal is $\alpha$.We have asked to calculate the velocity of the ball when the direction of motion of the ball makes an angle of $\beta$ with the horizontal.The horizontal component of velocity of the projectile remains the same throughout the motion of the projectile.Hence, the horizontal component of initial velocity ${v_{ix}}$ of the ball is
${v_{ix}} = v\cos \alpha$
Also, the vertical component of initial velocity ${v_{iy}}$ of the ball is
${v_{iy}} = v\sin \alpha$

Let us now determine the vertical component of velocity of the ball at time t.Rewrite equation (1) for the final vertical velocity of the ball.
${v_y} = {v_{iy}} - gt$
Substitute $v\sin \alpha$ for ${v_{iy}}$ in the above equation.
${v_y} = v\sin \alpha - gt$
The horizontal component of velocity of the ball remains the same.
${v_x} = v\cos \alpha$
Let us now write the expression for the angle between the components of velocity of the ball and velocity of the ball.
$\tan \beta = \dfrac{{{v_y}}}{{{v_x}}}$
Substitute $v\sin \alpha - gt$ for ${v_y}$ and $v\cos \alpha$ for ${v_x}$ in the above equation.
$\tan \beta = \dfrac{{v\sin \alpha - gt}}{{v\cos \alpha }}$
$\Rightarrow v\cos \alpha \tan \beta = v\sin \alpha - gt$
$\Rightarrow gt = v\sin \alpha - v\cos \alpha \tan \beta$
$\Rightarrow t = \dfrac{{v\sin \alpha - v\cos \alpha \tan \beta }}{g}$
$\Rightarrow g = \dfrac{{v\sin \alpha - v\cos \alpha \tan \beta }}{t}$

From the above equation, we can determine the components of the velocity of the ball when the angle made by the direction of motion of the ball with the horizontal is $\beta$.
${v_{fx}} = v\cos \alpha$
${v_{fy}} = v\cos \alpha \tan \beta$
Thus, the required velocity of the particle is
${v_f} = \sqrt {v_{fx}^2 + v_{fy}^2}$
$\Rightarrow {v_f} = \sqrt {{{\left( {v\cos \alpha } \right)}^2} + {{\left( {v\cos \alpha \tan \beta } \right)}^2}}$
$\Rightarrow {v_f} = \sqrt {{v^2}{{\cos }^2}\alpha + {v^2}{{\cos }^2}\alpha {{\tan }^2}\beta }$
$\Rightarrow {v_f} = \sqrt {{v^2}{{\cos }^2}\alpha \left( {1 + {{\tan }^2}\beta } \right)}$
$\Rightarrow {v_f} = \sqrt {{v^2}{{\cos }^2}\alpha {{\sec }^2}\beta }$
$\therefore {v_f} = v\cos \alpha \sec \beta$
Therefore, the required velocity of the particle is $v\cos \alpha \sec \beta$.

Hence, the correct option is B.

Note: The students may think that the horizontal component of velocity of the ball is taken the same for the initial angle of projection and final position of the ball. But the horizontal component of velocity of the ball does not change during its motion. The only changing component is the vertical component of the velocity.