The trajectory of a projectile in a vertical plane is y=ax-bx2,where a and b are constants,and x and y are respectievly the horizontal and the verticaldistanceof the projectile from the point of projection.The maximum height attained is?and the angle of projectionfrom the horizontal is?

Asked by Aditi Priya | 1st Oct, 2010, 08:33: AM

Expert Answer:

Dear Student,

The path of the projectile is a parabola. And the equation given is also of a parabola. But in simpler terms,

At maximum height, the path of the projectile is parallel to the ground, ie , only horizontal, this means at this point, the path is parallel to the x axis which means that the slope of the line at this point is zero. Now slope is given by differentiation.

So, differentiate the equation to get the slope

dy/dx= a-2bx

at max height , dy/dx= 0 =a-2bx hence x= a/2b
so, the value of y = 2aa/4b -aa/4b = aa/4b

hence the maximum height = aa/4b

Now for the angle of projection is the initial angle that the projectile makes with the x-axis. so the slope of the angle is given by dy/dx

hence tan(initial angle) = dy/dx at x=0

put x=0 in dy/dx above we get

tan(initial angle) = a

hence angle = tan inverse a

hence this is the angle of the projection.

We hope that clarifies your query.




Answered by  | 1st Oct, 2010, 03:44: PM

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