Answer:
The player won't score and will miss the basket for 2.07m short.
Explanation:
We have a 2D proyectile problem, the basketball is moving at the same time in the X and Y axes, we can start by writing the information given:
The distance between the player and the basket is 6.1m that will be X.
The height where the basket was launched is 1.2m that will be the initial position or [tex]Y_{0}[/tex].
The initial velocity, [tex]V_{o}[/tex] is 7.8 m/s and its launched within 55° with the horizontal, that will be α.
We also need the equations of 2D movemnt
[tex] 1. X=X_{0} +V_{x} *t\\2. Y=Y_{0} +V_{0Y} *t+\frac{1}{2}*a_{y} *t^{2}\\3. V_{x}=V_{0}*cos(\alpha)\\4. V_{y}=V_{0}*sin(\alpha)[/tex]
We can find the flying time t by using (1)
[tex] t=\frac{X-X_{0} }{V_{0}*cos(\alpha)} [/tex]
[tex] t=1.36s [/tex]
We can replace t in (2) and find what would be Y with that flying time, if it is equal to 3.05m the height of a basket the player would score, but if not he'll miss.
[tex]Y=0.83m[/tex]
As [tex]Y\neq 3.05m[/tex] the player won't score so we need to find how much will the ball miss the basket.
Using Y as 3.05m we can find the flying time necessary to score with equation 2.
we would obtain 2 answers for t, we have to chose the second one, remember that in a parabolic movement you pass 2 times through the same hight when you are going up and when you are going down so after finding the answers [tex]t=0.87s[/tex].
Finally replacing t found just before in (1) we can find the actual X, we will call it [tex]X_{2}[/tex] distance that the ball will travel.
[tex]X_{2}=3.93m[/tex]
and subtracting the two distances we will find by how much the ball will miss the basket.
[tex]X-X_{2}=2.07m[/tex]
So the player will miss the basket for 2.07 m.
