The problem: (Note: the correct answer is bolded throughout) A basketball is launched with an initial speed of 8 m/s at a 45 degree angle from the horizontal. The ball enters the basket in 0.96 seconds. What is the distance x and y? Just ignore the distance x, I know it's simply the product of the time and horizontal velocity. The distance y is obtained using the equation: y = (Vsin45)(t) + .5(-g)(t)^2 y = 0.91m The above is the correct answer and how to obtain it--no quarrels there. My issue is why this method does not work: Here are my variables: t = Total time the ball is in the air Tp = The time for the ball to reach its peak height Tf = The time the ball is in free fall Tp = (Vsin45) / (g) Tp = (8sin45) / (9.8) Tp = 0.57723s Tf = t - Tp Tf = 0.96 - 0.57723 Tf = 0.38277s y = .5(g)(t^2) y = .5(9.8)(0.38277^2) y = 0.72m The logic makes sense to me, since the ball takes .58 seconds to reach its peak height and obtain a velocity of 0 m/s. It should also then be in free fall for the remaining time (t-Tp) of 0.38s. I also thought that using the equation for displacement would be appropriate since only the acceleration due to gravity is present in the y direction. It is wrong however. If it helps, I'll also mention this problem was correctly solved using conservation of energy (0.91m), where the V in the final kinetic energy is obtained by using V horizontal, g, and the obtained Tf--so I know everything up until the displacement equation is correct. Could anyone enlighten me as to why I cannot use the displacement equation as I did?