Question

A particle is projected with speed $u$ at angle $\theta$ with vertical. Find maximum height.

• A. $H=\frac{u^2{\sin }^2\theta }{g}$
• B. $H=\frac{u^2{\cos }^2\theta }{g}$
• C. $H=\frac{u^2{\sin }^2\theta }{2g}$
• D. $H=\frac{u^2{\cos }^2\theta }{2g}$

Answer 1

The correct option is D $H=\frac{u^2{\cos }^2\theta }{2g}$

From figure : $u_x=usin\theta$ $u_y=ucos\theta$

Also $a_y=-g$ and $a_x=0$

y direction : $v_y^2-u_y^2=2a_y{S}_y$

$v_y=0$ at maximum height i.e $S_y=H$

$\therefore$ $0-(ucos\theta {)}^2=2(-g)H$ $⟹H=\frac{u^{2}co{s}^2\theta }{2g}$