(AP Physics) Can you please tell me how to figure this out? How do we find the initial velocity of a projectile that travels a distance of 52.5 meters and reaches a maximum height of 30 meters. How can I figure that out?

Question

anonymous · Answer

x= Vicos(theta)*t
y max = Visin(theta)*t/2 -1/2*g*(t/2)^2

anonymous · Answer

Sorry, you need a 3rd eqn.  :
0 = Visin(theta)*t -1/2*g*t^2

anonymous · Answer

ok?

anonymous · Answer

I don't have " t "

anonymous · Answer

nope.  but you have 3 equations with 3 unknowns.  you know what to do:)

anonymous · Answer

ok. And one more question.

anonymous · Answer

looking at these eqn.s, I see a little strategy for solving quickly...

anonymous · Answer

ok. how?
And thatnks for the help,

anonymous · Answer

x= Vxi*t  (no acceleration)
Vyf^2 = Viy^2 -2gh (solve at max height: Vyf =0)
use Vxi and Vyi  :    Vi =sqrt(Vxi^2+Vyi^2)

anonymous · Answer

x =52.5 and h=30

anonymous · Answer

make sense?

anonymous · Answer

I was solving and realized that they simplify into more compact equations...

anonymous · Answer

Yes. It makes a lot of sense now. Thank you. I appreciate the help.

anonymous · Answer

sure.

anonymous · Answer

oh yeah. And one more question. Since the projectile traveled a full length of 52.5, should I use 52.5 or just half of that?