Question

For a projectile fired at an angle 0 with the
horizontal, the maximum height is:​

Answer 1

if projectile angle is $\large\rm { \theta}$ from vertical, then

angle of projection $\large\rm { = 90 - \theta}$

Maximum height attained:

$\huge\rm { \frac{u^{2} \sin^{2} ( 90- \theta)}{2g} }$

$\huge\rm { = \frac{u^{2} \cos^{2} \theta }{2g}}$

$\large\rm { \therefore }$ Maximum height attained:

$\large\boxed{\sf { = \frac{u^{2} \cos^{2} \theta}{2g}}}$