Question

Derivation of equation of trajectory of a projectile

Answer 1

1) The path of the projectile is shown in the figure.
 

The horizontal component of velocity always remain same as there is no acceleration on the horizontal axis.

 

Thus, the horizontal distance travelled by the projectile is

 

x = u cosθ × t

 

t = x/(u cosθ)                                                   ...... (1)

 

The vertical distance travelled by projectile is given from kinematical equation s = ut + ½at2.

Here, s = y, a = ay = -g, u = uy = u sinθ

 

Thus, we have

 

y = u sinθ - ½gt2                                              ...... (2)

 

Substituting equation (1) in (2), we get
y=xtanθ−gx^2/2v^2cos^2(θ)

Where,
y is the horizontal component,
x is the vertical component,
g= gravity value,
v= initial velocity,
θ = angle of inclination of the initial velocity from horizontal axis,


 

Answer 2

Answer:

Answer is given below.