Please solve this physics question for me.......

Question

A projectile is fired in such a way that its horizontal range is equal to three times its maximum height.What is the angle of projection?

Answer 1

Let:

x be the horizontal distance at time t,

v be the vertical speed at time t,

y be the vertical height at time t,

g be the acceleration due to gravity,

a be the angle of projection measured from the horizontal.

x = ut cos(a) ...(1)

y = ut sin(a) - gt^2 / 2 ...(2)

v = u sin(a) - gt ...(3)

y is maximum when v = 0.

Putting v = 0 in (3) gives t = u sin(a) / g.

Substituting this value of t in (2) gives:

y = u^2 sin^2(a) / (2g) ...(4)

From (2), when the projectile is at ground level:

0 = t(u sin(a) - gt / 2)

This is initially when t = 0, and on return when:

t = 2u sin(a) / g.

Substituting this value of t in (1):

x = 2u^2 sin(a)cos(a) / g ...(5)

From (4) and (5), the ratio of range to maximum height is:

x / y = 4 / tan(a)

Putting x / y = 3 gives:

tan(a) = 4 / 3

a = 53.1 deg.