Question

Equation of trajectory for a projectile is y = 2x{1-(x/40)} (where x and y are in meter). What will be the range of the projectile?

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 `alpha` with the horizontal direction (x-axis).
The vertical component of the velocity (along y-axis) of projection is `usinalpha` and the horizontal component is `ucosalpha`

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=usinalphaxxT+1/2gT^2`
`=>0=uxxT-1/2xxgxxT^2`
where `g="acceleration due to gravity"`
`:.T=(2usinalpha)/g`
The horizontal displacement during this T sec or the Range ,`R=ucosalpha xxT`

`=>R=(2u^2sinalphacosalpha)/g...(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=ucosalphaxxt....(2)`

And the vertical displacement during t sec

`y=usinalphaxxt-1/2xxgxxt^2.....(3)`

Combining (1) and (2) we get

`y=usinalphaxxx/(ucosalpha)-(gx^2)/(2u^2cos^2alpha)`

`=>y=xtanalpha-x^2/((2u^2cos^2alpha)/g)....(4)`

This is the equation of the trajectory,we obtained.

Now in the given problem the equation of the trajectory is

`y=2x(1-x/40)`

`=>y=2x-x^2/20......(5)`

Comparing (4) and (5) we get

`tanalpha=2.....(6)`

and

`(2u^2cos^2alpha)/g=20.....(7)`

Multiplying (6) by (7) we get

`(tanalphaxx2u^2cos^2alpha)/g=2xx20`

`=>(2u^2sinalphacosalpha)/g=40`

`=>R=40`

So Range of the projectile `R=40m`