Recent content by rosekaty

Yes I got 63°

Thank you all for your help ! Really I mean it, it wasn't easy for me and you helped me till the end. Double difficulty I'm French :wink:

I GET IT ! I was in radian !

Lol Yes I checked many times I'm pretty sure i's false. I did arctangent(19.79/10) because we found Vy=19.44 and Vx=10 m*s^-1

Yes thank you I just get that ! It was simple. I found an angle of 1° :confused:

Well...I tried it and for the first equation I realize that I don't have the maximal height since I don't have the height of the building. And the second one seems..unsolvable.

There is a lot of equations involivng the angle, I could pick the one with the maximum height that says: H= (initial velocity^2*sin(alpha)^2)/2*g Or tan(alpha) = Vo*sin(alpha)/Vo*cos(alpha)

No, so if I use what you told me earlier Vo=22.18 m*s^-1 :smile:

Okay so: final velocity^2=Initial velocity^2+2*a*H 0=Initial velocity+2*(-9.8)*20 Initial velocity = 19.79 m*s^-1

Vertical velocity= -g*t+ maximum height ? or (y(x)-L-(0.5*g*t^2))/t=Vertical velocity (To be honest, I think I'm lost. I see your point but there is so many equations I don't know which one I should use.)

Well I think the maximum height can be useful. Maybe if I use this equation: the maximum height =(vo^2*sin^2(alpha))/2*g

The horizontal speed is 10 m*s^-1 then.

Because of the total height of the building that I don't have, I don't think I can. If I "cut" the motion to the top of the building to a point on the motion in the same horizontal, Delta x still missing even if I have Delta y which is the 20meters.

Yes, I thought about that and I got: the launch point = x= -L or x=Vo^2*((2*L*cos^2*alpha+sin(2*alpha)))/g the maximum height =(vo^2*sin^2(alpha))/2*g

Homework Statement A ball is thrown at the edge of the roof of a building in a direction at an "alpha" angle above the horizontal. It landed 5 seconds later at 50 meters from the building. The maximum height reached by the ball during its trajectory is 20 meters above the roof. Find The...