LoungeFor casual talk about things unrelated to General Motors. In other words, off-topic stuff. And anything else that does not fit Section Description.

Hey guys, I'm in physics (high school still but I graduate from HS this year yeah baby!) anyway, I'm stuck on a problem and I thought "maybe the good fellas from BC can help me"

So here it is.

A projectile is fired at a speed of 20 meters/sec and at an angle of 35 degrees. What is the max height the projectile will achieve?

I think you use this equation: y = vy (t) + (1/2)at^2
y is y position, vy is initial y direction velocity, a is gravity, and t is time
First get the inital velocity component at 35*. Using that right hand triangle stuff SOHCATOA, the y component at 35* is (20 m/*) x sin (35*). That'* your vy.
Now I think you put 0 for y since that'* where it'* going to end up (at the same height).
Gravity is -9.81 m/*^2.
Substitute everything in, solve for t.
Then take that t, divide it by 2 since the max height will be in the middle of flight.
Put that t/2 back into the original equation and solve for y.
I think thats how you do it but I haven't done a problem like that in 3 years..

DUDE, That'* exactly right!!! Thank you so much, I was struggling for a long time, I couldn't remember how I did it. It'* review for my midterms.
I must have forgotten about the solving for time.
You're a lot nicer than Hans.

Thanks again, now I know who to ask for help next time.