Question

A ball is thrown upward with velocity $v$ and at an angle $\theta$ with the ground. What will be velocity of the ball after $t$ seconds? (Given mass of the ball $=m$ )

• A. $\sqrt{{\left(vcos\theta \right)}^2+{\left(vsin\theta \right)}^2}$
• B. $\sqrt{{\left(vcos\theta -vsin\theta \right)}^2-gt}$
• C. $\sqrt{v^2+g^2{t}^2-\left(2vsin\theta \right)gt}$
• D. $\sqrt{v^2+g^2{t}^2-\left(2vsin\theta \right)gt}$

Answer 1

Answer: (C)

As we know that for a projectile motion the instantaneous velocity after $tsec$ is given by:
$\left|\overrightarrow{V_t}\right|=\sqrt{v_x^2+v_y^2}$
Where $v_x=vcos\theta$ is the horizontal component of velocity.
$v_y=vsin\theta -gt$ is the vertical component of velocity .
Hence putting these values in the above equation,we get:
$\Rightarrow \left|v_t\right|=\sqrt{{\left(vcos\theta \right)}^2+{\left(vsin\theta -gt\right)}^2}$
$\Rightarrow \left|v_t\right|=\sqrt{v^2+g^2{t}^2-2vsin\theta gt}$