Question

what are the two angles of projection of a projectile projected with velocity 30 metre per second so that the horizontal range is 45 m
(take g=10 m ^-2) ​

Answer 1

Answer:

$range = \frac{v {}^{2}sin (2a) }{9} \\ = > 45 = \frac{900 \sin(2a) }{10} \\ = > 45 = 90sin(2a) \\ = > sin(2a) = \frac{1}{2} \\ = > 2a = 30degree \: or \: 150degree \\ = > a = 15degree \: or \: 75degree$

Answer 2

Answer:

,this is the answer..............