Question

For a projectile launched from ground at an angle 0 with the vertical, the maximum height reached is directly proportional to sin-0 directly proportional to-cos²0 directly proportional to 1/sine directly proportional to 1/sin-0​

Answer 1

Answer:

Explanation:

Angle of projection is taken from horizontal direction.

So, angle of projection θ=90−β

Maximum height reached H=

2g

u

2

sin

2

θ

​

=

2g

u

2

cos

2

β

​

⟹ ucosβ=

2gH

​

Time to reach maximum height T=

g

usinθ

​

Or T=

g

2sin(90−β)

​

=

g

ucosβ

​

=

g

2H

​