4.32 (a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by

θ(t)=tan1v0ygtv0x

(b) Shows that the projection angle θ0 for a projectile launched from the origin is given by

θ0=tan14henR

where the symbols have their usual meaning.

NEETprep Answer:
 
Let v0x and v0y respectively be the initial components of the velocity of the projectile along horizontal (x) and vertical (y) directions.
 
Let vx and vy respectively be the horizontal and vertical components of velocity at a point P.
 
 
Time is taken by the projectile to reach point P = t
Applying the first equation of motion along with the vertical and horizontal directions, we get:
 
vy=v0y=gt And vx=v0xtanθ=vyvx=v0ygtv0xθ=tan1(v0ygtv0x)
Maximum vertical height hm=u02sin2 2θ2g                             ...i
 
Horizontal range,R=u02sin2 2θg                           ...ii
Solving equations (i) and (ii), we get:
 
hmR=sin2 θ2sin2 θ          =sin θ×sin θ2×2sin θcos θ          =14sin θcos θ=14tan θtan θ=(4hmR)θ=tan1 (4hmR)