Question

A body of mass 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. $\sqrt{(vcos\theta +(v\sin \theta {)}^2}-gt$
• 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 $\sqrt{v^2+g^2{t}^2-\left(2vsin\theta \right)gt}$

In order to find velocity separately horizontal and vertical component then use vector to find final velocity.

We have

$\text{so,}{v}_x=u_x+a_{x}t{v}_x=ucos\theta +0.t{v}_x=ucos\theta$

$\text{similarly,}{v}_y=u_y+a_y.t{v}_y=usin\theta +(-g)t{v}_y=usin\theta -gt$

now, $v=v_x\hat{i}+v_y\hat{j}$

so,$v=ucos\theta \hat{i}+(usin\theta -gt)\hat{j}$

now magnitude of velocity, |v| = $\sqrt{(ucos\theta {)}^2+(usin\theta -gt{)}^2}$

= $\sqrt{u^{2}co{s}^2\theta +u^{2}si{n}^2\theta +g^2{t}^2-2usin\theta gt}$

= $\sqrt{u^2(co{s}^2\theta +si{n}^2\theta )+g^2{t}^2-2usin\theta gt}$

= $\sqrt{u^2+g^2{t}^2-2usin\theta gt}$