The equation of motion of a projectile is `y = 12 x - (3)/(4) x^2`. The horizontal component of velocity is `3 ms^-1`. What is the range of the projectile ?

Options

A. `18m`
B. `16m`
C. `30.6m`
D. `21.6m`

Answer

B

Explanation

`y=12x-3/4x^(2)`
`(dy)/(dt)=12 (dx)/(dt)-3/2x(dx)/(dt)`
At x=0, `(dy)/(dt)=12 (dx)/(dt)`
If `theta` be the angle of projection, then
`(dy//dt)/(dx//dt)=12 = tan theta`
Also, if u=initial velocity, then `ucos theta=3`
Hence, `tan thetaxxucos theta=36 or usin theta=36`
Range `R=(u^(2) sin 2theta)/g=(2u^(2)sin theta cos theta)/g`
`=(2(u sin theta)(u cos theta))/10=(2xx36xx3)/10=21.6m`