In other words, m = (v0 cos 45} t Let's now write down a second equation. this time for the y-direction. Y = VOY'E + 3/2 :3le2 Equation 2 {Remember that ay = -g = -9.80 m/s Since the target was deliberately chosen to be the same height as the launch, y equals 0; divide the remaining terms by t , rewrite the equation, spelling out V0}? as V0 51119 O = v0 sin 45" + 1/: ayt Recall that sin45 = cos 45 = 0.707. Rewrite your equations for the x-direction and y-direction with this number plugged in: Note that you now have two equations in two unknowns namely, V and E . However, 0 there is no point in determining the time t as you have no way of measuring it. Eliminate t by solving both equations for t and setting the two expressions equal to each other: You now have one equation in one unknown, namely V j the ball's actual speed as it left the D launcher. Solve for it