The hint my prof gave me (which doesn't make sense to me) was g) subs half of the literal solution for t from part e) into the y(t) fxn and simplify. h) Solve the x(t) fxn for t literally and subs that into the y(t) fxn and simplify. Another hint I found somewhere else was: g)Plug in 270 for x in the x = ... equation, then solve for t
h) Replace sec^2 with 1/(cos^2) and tan with (sin/cos) and see if you get the original equations.

I also found out that the projectile is in the air for about 5.4 seconds. The projectile lands 270 feet from the gun. And that the max height is 117.2 feet. So I hope some of that info helps in solving. The solution from part e) was t = V0sinT/16. If anyone can help me, I will be forever thankful! Thank you in advance!

Okay so am I on the right track for part g)... is the answer either 270 = V0(5.4s)cosT = 50 = V0*cosT OR 270 = V0(2.7s)cosT = 100 = V0*cosT?

for h) if I sub (1/cos^2T) and (sinT/cosT) in for tanT where do I go from there?

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