Question

A body of mass m is thrown upwards at an angle θ with the horizontal with velocity (v) . While rising up

the velocity of the mass after t seconds will be

Answer 1

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

explanation : it seems we have to use $\textbf{projectile motion}$ concepts .

here, $u_x=ucos\theta, a_x=0$

$u_y=usin\theta, a_y=-g$

we have to find velocity after t sec .

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

$\text{so,}\: v_x=u_x+a_xt\\v_x=ucos\theta+0.t\\v_x=ucos\theta$

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

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| = $\sqrt{(ucos\theta)^2+(usin\theta-gt)^2}$

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

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

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

Answer 2

A body of mass m is thrown upwards at an angle θ with the horizontal with velocity (v) . While rising up

the velocity of the mass after t seconds will be √v²+g²t²−(2vsinθ)gt.

Explanation:

Instantaneous velocity of rising mass after t sec will be √vt=v²x+v²y where vₓ=vcosθ=Horizontal component of velocity v=vsinθ−gt=Vertical component of velocity vt=√(vcosθ)2+(vsinθ−gt)

2=√v²+g²t²−2vsinθgt