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)

 

[Nickie]

I can understand your confusion about why the displacement equation did not work in this scenario. However, the displacement equation you used, y = .5(g)(t^2), assumes that the initial velocity is 0 m/s. In this problem, the initial velocity is 8 m/s at a 45 degree angle, which means that the ball already has a vertical component of velocity when it starts its journey. This initial vertical velocity must be taken into account when calculating the displacement.

The equation y = (Vsin45)(t) + .5(-g)(t)^2 takes into account both the initial vertical velocity (Vsin45) and the acceleration due to gravity (-g) to accurately calculate the displacement. This equation is derived from the kinematic equations, which take into account the initial velocity, acceleration, and time to calculate displacement.

In summary, the displacement equation you used only works when the initial velocity is 0 m/s. In this problem, the initial velocity is not 0 m/s, so it cannot be used to accurately calculate the displacement. I hope this explanation helps to clarify the issue.