Question

A particle is projected with speed $v_0$ at angle $\theta$ above the horizontal such that the ratio of kinetic energy at highest point to that at the point of projection is $3:4$, then change in velocity of the particle between point of projection and the highest point will be.

• A. $\frac{v_0}{2}$ vertically downward
• B. $\frac{v_0}{2}$ vertically upwards
• C. $2v_0\sqrt{3}$ due east
• D. $\frac{v_0\sqrt{3}}{2}$ vertically downward

Answer 1

The correct option is A $\frac{v_0}{2}$ vertically downward

$V_{highest}=V_0\cos \theta K_h=\frac{1}{2}m(V_0\cos \theta {)}^2{K}_i=\frac{1}{2}(V_0{)}^2\frac{3}{4}={\cos }^2\theta \theta =30°$

So, change in velocity in upward direction $=0-V_0\sin 30°$

$=\frac{V_0}{2}$ vertically downward