Question

If a projectile is fired at an angle $\theta$ with the vertical with velocity $u$, then maximum height attained is given by:

• A. $\frac{u^{2}cos\theta }{2g}$
• B. $\frac{u^{2}si{n}^2\theta }{2g}$
• C. $\frac{u^{2}si{n}^2\theta }{g}$
• D. $\frac{u^{2}co{s}^2\theta }{2g}$

Answer 1

The correct option is D $\frac{u^{2}co{s}^2\theta }{2g}$

If projection angle is $\theta$ from vertical then, angle of projection is $90-\theta$.
maximum height attained $=\frac{u^{2}si{n}^2(90-\theta )}{2g}=\frac{u^{2}co{s}^2\theta }{2g}$