Question

Find the angle of projection for which maximum height of projectile is 1 /3rd of
it's horizontal range.
Solution:​

Answer 1

Explanation:

ANSWER

R=

g

u

2

sin2θ

H=

2g

u

2

sin

2

θ

g

u

2

sin2θ

=

2g

u

2

sin

2

θ

2sinθcosθ=

2

sin

2

θ

ucosθ=sinθ

tanθ=4

θ=tan

−1