Class 11 NEET Physics Motion in a Plane Questions

Let ϕ be elevation angle of the projectile at its highest point as seen from the point of projection O and θ be angle of projection with the horizontal. From figure,
tan‌ϕ=

H

R/2

............(i)
In case of projectile motion
Maximum height, H=

u2sin2θ

2g

Horizontal range,R =

u2sin2θ

g

Substituting these values of H and R in (i), we get
tan‌ϕ=

sin2θ

sin‌2‌θ

=

sin2θ

2‌sin‌θ‌c‌o‌s‌θ

=

1

2

tanθ
Here, θ=45°
∴tan‌ϕ=

1

2

‌tan‌45°=

1

2

;ϕ=tan−1(

1

2

)