Question

A body is projected with velocity 60 metre per second at angle 30degree with horizontal what is maximum height attained​

Answer 1

Answer:

$maximum \: height \: = \frac{ {u}^{2} \sin( \alpha ) }{2g} \\ h \: = \frac{ {60}^{2} \times \frac{1}{4} }{20} \\ h \: = 45$

Explanation:

hope it will help you