Recent content by Francis Perez

Thank you everyone, I found the formula :)

Well, I was doing trial and error. It seems that I should use Vy=Vi(sin(Theta)) from the very beginning. Vi is correct, (Vy2+Vx2) So I did and I got Arcsin(Vy/Vi) = theta Then I did Xf*Tan(Theta) to get the the point at which my projectile should aim and add that to the Yf.I think I was...

Yo have misunderstood me, it is not dierected exactly at the point where target (Xf,Yf) is located at. It’s actually getting the angle that is shooting to get to the Max height. Then using that angle to find the height above the Y value that the projectile must aim at in order to arc down to the...

Oooh I’m dumb.. so the equation ##t_{FromMaxHToY}## would be sqrt(5*2)/g Then.. I was thinking instead of using the angles they shoot at. I should instead find the Time when the projectile reaches Yf+5. To do this I have tried finding X-Magnitude from Xi to Xf. Then I tried using the time from...

So I’m guessing: Vy= sqrt(2*Grav*(Yf+5)) For Y(t)Max = Yf +5 Yf+5= Vy(t) - (1/2)g*t2 Or Yf+5 = (sqrt(2*Grav*(Yf+5)))*t - (1/2)g*t2 I’m hoping this is going to work no matter where the Target, (Xf[/SUB, Yf), are at. The projectile will do an arc 5 meters above the Yf value before going down to...

So let’s see.. tMaxHeight = Vy/g.. Then I find with this tFromMaxHToY =sqrt( ((Y+5)*2)/g) Then.. tToPointY =tMH+tMHtoY And now at this time, I have to use the X and the time to find the Vx, right? So Vx = X/t? After that.. how do we find the optimal angle and initial velocity for these points...

Homework Statement Hello, so I’ve been trying to find some possible way to calculate this: The initial Velocity and angle needed to launch a projectile and reach the coordinates (x,y). The maximum height the ball can reach before it starts to head down to the target value should be y+5m The...