Question

If the trajectory followed by the projectile projected from the ground is y = x , where x and y are in metre, then the range of projectile is [here x-axis is taken along horizontal and y-axis is taken along vertical

Answer 1

Let the velocity of projection of the projectile in x-y plane from the origin (0,0) be u with angle of projection α with the horizontal direction (x-axis).

The vertical component of the velocity (along y-axis) of projection is usinα and the horizontal component is ucosα

Now if the time of flight be T then the object will return to the ground after T sec and during this T sec its total vertical displacement h will be zero. So applying the equation of motion under gravity we can write

h=usinα×T+12gT2

⇒0=u×T−12×g×T2 

where g=acceleration due to gravity

∴T=2usinαg

The horizontal displacement during this T sec or the Range ,R=ucosα×T

⇒R=2u2sinαcosαg...(1)

Let the position of the projectile in x-y plane after t sec of its projection be (x,y)

The horizontal displacement during t sec

x=ucosα×t....(2)

And the vertical displacement during t sec

y=usinα×t−12×g×t2.....(3)

Combining (1) and (2) we get

y=usinα×xucosα−gx22u2cos2α

⇒y=x