A projectile is fired in such a way that its horizontal range is equal to 11 times its maximum height.?

Question

What is the angle of projection?

Answer 1

The trajectory is a parabola, and an inverted one at that. It has a zero at the point of firing, a zero where the projectile lands and is positive in between. This has an equation of the form:
y = f(x) = ax(r-x) = (ar)x - ax^2

where a is a positive constant and r is the horizontal range. Verify that this is quadratic and meets the conditions f(0)=f(r)=0 and f(x)>0 for [0 f'(x) = ar - 2ax
f'(0) = ar

So arctan(ar) is the answer...but we don't know what a and r are. We do know that the peak altitude happens at the midpoint (x=r/2), and we know the value of f there is one 11th of the range, which gives:
f(r/2) = r/11
a*(r/2)(r-r/2) = r/11
a*(r/2)*(r/2) = r/11
ar = 4/11

So, the angle is arctan(4/11) ... just a hair under 20 degrees.

Answer 2

R = 2 H(max) u^2 sin 2p / g = 2 [u^2 sin^2 p] / 2g sin 2p = sin^2 p 2 sin p cos p - sin^2 p = 0 sin p [2 cos p - sin p] = 0 sin p = can't be 0 because it makes p = 0 -------------------------------- 2 cos p = sin p tan p = 2 p = sixty 3.40 3 degree above horizontal