The equation of the trajectory of a projectile on a vertical plane is y=`ax-bx^2`, where a and b are constants, and x and y respectively are the horizontal distances of the projectile from the point of projection. Find out the maximum height attained by the projectile, and the angle of projection with respect to the horizontal.

Explanation

Let ,u= velocity of projection ,`alpha` =angle of projection .
The velocity u cos `alpha` in the horizontal direction is uniform. So, in time t,
`x= u cos alpha *t or, t=(x)/(u cos alpha)`
The velocity u sin `alpha` in the vertically upward direction is under a uniform retardation -g , where g is the acceleration due to gravity. Then, in time t,
`y= u sin alpha*t-1/2 "gt"^2=u sin alpha*(x)/(u cos alpha)-1/2g""(x^2)/(u^2 cos^2alpha) or, y=x tan alpha-(g)/(2u^2 cos^2alpha)x^2`
Comparing with the given equation =`ax -bx^2`, we get
(i) a=tan`theta` or, angle of projection , `theta =tan^(-1)a`.
(ii) b=`(g)/(2u^2 cos^2 alpha) or, u^2(g)/(2b cos^2 alpha)`
At maximum height H, the velocity of the projectile is zero . Considering vetical motion, we have
0=`(usin alpha)^2 -2gH`
or, `H=(u^2sin^2alpha)/(2g)=(g)/(2bcos^2alpha)*(sin^2alpha)/(2g)`
`=(tan^2alpha)/(4b)=(a^2)/(4b)`.