Question

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

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

Answer 1

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

Instantaeous speed of rising mass after t sec will be $v_t$ = $\sqrt{{v_x}^2+{v_y}^2}$
where $v_x$ = v sin $\theta$ = Horizontal component of velocity
$v_y$ = v sin $\theta$ - gt = Vertical component of velocity
$v_t$ = $\sqrt{{\left(vcos\theta \right)}^2+{(vsin\theta -gt)}^2}$