First slide

Projection Under uniform Acceleration

Question

A particle is projected from the ground with velocity u making an angle θ with the horizontal. At half of its maximum height,

Difficult

A

its horizontal velocity is ucosθ
B

its vertical velocity is usinθ / √2
C

its velocity is $\large u(\frac {1+cos^2\theta}{2})^{1/2}$
D

all the above are true

Solution

$\large {\nu _x} = {u_x} = u\cos \theta \;;\;{\nu _y} = \sqrt {{u_y}^2 - 2gy}$
$\large {\nu _y} = \sqrt {{u^2}{{\sin }^2}\theta - \frac{{{u^2}{{\sin }^2}\theta }}{2}} = \frac{{u\sin \theta }}{{\sqrt 2 }}$
$\fn_cm \large v = \sqrt{v_x^2+v_y^2}=\sqrt{(ucos\theta)^2+(\frac{usin\theta}{\sqrt2})^2}$
$\large = u\sqrt {\frac{{2{{\cos }^2}\theta + {{\sin }^2}\theta }}{2}} = u\sqrt {\frac{{1 + {{\cos }^2}\theta }}{2}}$

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