Question

A body of mass m is thrown upwards at an angle $\theta$ with the horizontal with velocity v. While rising up the velocity of the mass after t seconds will be

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

Answer 1

The correct option is C $\sqrt{v^2+g^2{t}^2-\left(2vsin\theta \right)gt}$

Instantaneous velocity of rising mass after t sec will be
$v_t=\sqrt{v_x^2+v_y^2}$
Where $v_x=vcos\theta$ =Horizontal component of velocity
$v_y=vsin\theta -gt=Verticalcomponentofvelocity$
$v_t=\sqrt{(vcos\theta {)}^2+(vsin\theta -gt{)}^2}$
$v_t=\sqrt{v^2+g^2{t}^2-2vsin\theta gt}$