- #1
Greyt
- 4
- 0
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?
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?