First slide

Projection Under uniform Acceleration

Question

A particle is projected with a speed v₀ at an angle $θ$ above the horizontal surface such that the ratio of the kinetic energy at the highest point and the point of projection is 3:4. The change in velocity of the particle between these two points is

Moderate

A

(v₀/2) vertically downward.
B

(v₀/2) vertically upward.
C

$\sqrt{3} v_{0}$ due east.
D

$(\sqrt{3} v_{0} / 2)$ vertically downward.

Solution

The ratio of the kinetic energy at the highest point and point of projection is
$\begin{array}{l} \frac{K_{H}}{K_{0}} = \frac{3}{4} \\ \Rightarrow \frac{(1 / 2) m {(v_{0} \cos θ)}^{2}}{(1 / 2) {mv}_{0}^{2}} = \frac{3}{4} \\ \Rightarrow \cos^{2} θ = \frac{3}{4} \Rightarrow θ = \frac{π}{6} \end{array}$
Change in velocity $= v_{0} sinθ = v_{0} / 2$ .

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