hard Ap physics question?

Question

An astronaut on the Moon fires a projectile from a launcher on a level surface so as to get the maximum range. If the launcher gives the projectile a muzzle velocity of 26 m/s, what is the range of the projectile? [Hint: The acceleration due to gravity on the Moon is only one sixth of that on the Earth.]

Answer 1

Range=(Vo^2/a)*sin(2theta)

The a in this case is the magnitude of the acceleration. The maximum range is acheived at an angle of 45 degrees.

Answer 2

d= 413.9 meters

Vx = Velocity in x direction = V *Cos(Angle)
Vy= Velocity y direction (up)= V*Sin (Angle)
G=(9.8m/sec^2)/6 = 1.6333 m/sec^2
T = time of flight =2*Time to reach Max elevation

T=2*Vy/G

d= T*Vx

Therefore

d=2*VyVx/g
=2*V^2*Cos(Angle)*Sin(Angle)/g

I'll let you prove that Cos*Sin has a maximum at 45 degrees (which it does).

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Since I started figuring the answer, Science Geek looked up the formula.

Note: Cos(theta)*Sin(theta) = Sin (2theta)/2 so our answers are the same, except that it is much easier to prove Sin (2theta) has a max at 45 degrees.