2D kinematics - Solved, but a question on the logic

[Greyt]

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?

 

[haruspex]

Think about what that y you computed as 0.72 actually represents.

 

[SammyS]

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?

The value you get for y, 0.72m is the y value from the ball's peak height to the basket height. It should actually be -0.72m.

What is the height gained by the ball in its first 0.57723 of flight?

 

[Greyt]

Ah, I completely neglected the fact it was traveling down from its peak and so the y represents how far the basket is from the top of its height.

Thanks to you both! (I feel silly for ignoring that negative sign for gravity now)