Question

Hello, my cat has to jump to a platform at coordinate (Xa,Ya) with a jump speed of "JumpSpeed".
He is currently at coordinate (0,0). And "t" = 0, when he jump.
What Angle "a" should he jump ? (note It's note necessary possible)

JumpVelX = JumpSpeed * cos(a)
JumpVelY = JumpSpeed * sin(a)

Position = t*V + (1/2)t²*g

g = gravity (0,-9.81)

positionX = t * JumpSpeed * cos(a)
positionY = t * JumpSpeed * sin(a) + (1/2)t² * g

with positionX we get :
t = positionX/(cos(a)*JumpSpeed)
....

Answer 1

lets replace positionY's "t" with it :
positionY = t * JumpSpeed * sin(a) + (1/2)t² * g
positionY = (positionX/(cos(a)*JumpSpeed)) * JumpSpeed * sin(a) + (1/2)(positionX/(cos(a)*JumpSpeed))² * g

But i can't resolver "a" (angle).

Answer 2

\[PositionY = \frac{ PositionX }{ \cos(a) * JumpSpeed } * JumpSpeed * \sin(a) + \frac{ 1 }{ 2 } * \frac{ PositionX }{ \cos(a) * JumpSpeed } * g\]

Answer 3

http://en.wikipedia.org/wiki/Trajectory_of_a_projectile#Angle_required_to_hit_coordinate_.28x.2Cy.29 got it