Question

In case of
projectile motion if the maximum height reached is1/4th of the horizontal range,the angle of projection is

Answer 1

Answer: 45 degree.

Explanation:

Maximum height, H = (u^2sin^2￥)/2g

Horizontal range, R = (u^2sin2￥)/g

According to question, H = R/4

4H = R

4 (u^2sin^2￥)/2g = (u^2sin2￥)/g

2u^2sin^2￥= u^2sin2￥

2sin^2￥= 2sin￥cos￥

sin￥ = cos￥

tan￥ = 1

￥= 45

Answer 2

Answer: Angle of projection = 45°

Explanation:

Taking maximum height = Hmax

Horizontal Range = R

Given : Hmax = 1/4 × R .....(1 )

Formula :

1) Height in projectile at angle

= u²sin²@/ 2g

2) Horizontal Range = u²sin2@/g

Taking , Hmax= u²sin²@/2g

 R = u²sin2@/g

substituting the above two formulas in equation (1)

u²sin²@/ 2g = 1/4× u²sin2@/g

solving the above equation ,

2sin²@ = sin2@

since , sin2@ = 2sin@cos@

therefore ,

2sin²@ = 2sin@cos@

or

sin@ = cos@

or

tan@ =

@ = tan^-1(1)

@ = 45°