Question

Derive an equation of the trajectory of a projectile and hence show that that the trajectory is a parabola​

Answer 1

$\textbf{Answer is in attachment!}$

Answer 2

Explanation:

In order to derive the equation of trajectory for a projectile particle, we in other words consider the relation between the x and yth co-ordinate.

Now, we know that for;

               x = u cos ∅t

               x/ u cos ∅ = t _______(1)

Also,        y = (u sin ∅)t - 1/2 gt² ______(2)

On substituting the value of (1) in (2) , we get;-

               y = (u sin ∅)x/ u cos ∅ - 1 /2 g × x²/ u²cos²∅  

               y = x tan ∅ - gx² / 2 u²cos²∅

               y = ax - bx²  (let tan∅= a & g/2u²cos²∅= b)

Clearly, the equation denotes parabolic origins. Hence, the path of trajectory will be parabolic.

Then,  

               y = x tan ∅ - gx² / 2 u²cos²∅

               y = x tan ∅ - gx² / 2 u²cos²∅ × tan ∅/tan ∅  

               y = x tan ∅ [ 1 - gx/ 2 u²cos²∅ × tan ∅ ]

               y = x tan ∅ [ 1 - gx/ 2u² cos²∅ × sin∅/cos∅]

               y = x tan ∅ [ 1 - gx/ 2u² cos∅ × sin∅]

               y = x tan ∅ [ 1 - x/ R ]    (since, g/ 2u²sin∅×cos∅= 1/R from the formula of horizontal range, in case, here given is the inverse of it).

Hence,

             y = x tan ∅ [ 1 - x/ R ]