We need to consider a right angled triangle with one angle α for the trajectory.

cos α = adjacent / hypotenuse
Adjacent = cos α × hypotenuse


Horizontal distance travelled , x

= Adjacent

= cos α × hypotenuse

= cos α × (Speed × time)

(The hypotenuse is basically the distance travelled by the particle if it was not affected by gravity and air resistance. It would go in the direction of the elevation unchanged)

cosα × u × t

(u is speed → distance/time and t is time)

= ut cos α

To account for gravity we have to consider the acceleration due to it.

Basically integrating acceleration once gives us velocity. Integrating velocity gives us displacement.

g is a constant and we can take it to be 10m/s²)

∫g dt = gt + c

We know that c = 0 since when t = 0, the particle has no velocity. It hasn't been fired yet.


∫gt dt = ½gt² + d

d = 0 since we know we started from the horizontal, meaning it is 0m above ground at t = 0.

so the distance it should drop due to gravity = ½gt²

Then we subtract the distance that the particle would fall by due to gravity from the distance travelled upwards to get the overall displacement.)

Assume g = 10m/s²





Vertical distance travelled , y

= Distance of the opposite side - ½gt²

= sinα × hypotenuse

= sinα × u × t - ½(10)t²
= ut sinα - 5t²

Then,

x² + y²


= (ut cosα)² + (ut sinα - 5t²)²

= u²t² cos²α + (ut sinα)² - 2(ut sinα)(5t²) + (5t²)²

= u²t² cos²α + u²t² sin²α - 10ut³ sinα + 25t⁴

= u²t² (cos²α + sin²α) - 10ut³ sinα + 50t⁴ - 25t⁴

= u²t² (1) - 10t²(ut sinα - 5t²) - 25t⁴

(Recall that cos²θ + sin²θ = 1)

= u²t² - 10t²y - 25t⁴

= u²t² - 10yt² - 25t⁴

(Shown)

Edited. Comment if you have any questions about the working.

Some useful physics formulas to recall for such questions :

Displacement , s = ut + ½ at² , where t is time, a is acceleration and u is initial velocity

Velocity at any time, v = u + at


ds/dt = v
dv/dt = a


Acceleration is taken to be constant



For this question, a = g = 10m/s²




Kinetic Energy = ½mv², where m = mass and v is velocity