Question

A basketball leaves a player's hands at a height of 2.20 m above the floor. The basket is 2.70 m above the floor. The player likes to shoot the ball at a 36.0 ∘ angle. Of the shot is made from a horizontal distance of 9.10 m and must be accurate to ±0.23m (horizontally), what is the range of initial speeds allowed to make the basket

Answer 1

The range of initial speeds allowed to make the basket is:
`9.954\,(m)/(s)\leq v \leq 10.185\,(m)/(s)`.

Step-by-step explanation:

We must notice that basketball depicts a parabolic motion, which consists of combining a constant speed motion in x-direction and free fall motion in the y-direction. The motion is described by the following kinematic formulas:

x-Direction

`x = x_(o) + v_(o)\cdot t \cdot \cos \alpha`

y-Direction

`y = y_(o) + v_(o)\cdot t\cdot \sin \alpha +(1)/(2)\cdot g\cdot t^(2)`

Where:

`x_(o)`,
`y_(o)` - Initial position of the basketball, measured in meters.

`x`,
`y` - Final position of the basketball, measured in meters.

`v_(o)` - Initial speed of the basketball, measured in meters per second.

`t` - Time, measured in seconds.

`\alpha` - Tilt angle, measured in sexagesimal degrees.

`g` - Gravitational acceleration, measured in meters per square second.

If we know that
`x_(o) = 0\,m`,
`y_(o) = 2.20\,m`,
`\alpha = 36^(\circ)`,
`g = -9.807\,(m)/(s^(2))`,
`x = (9.10\pm0.23)\,m` and
`y = 2.70\,m`, the system of equation is reduce to this:

`(9.10\pm 0.23)\,m = 0\,m + v_(o)\cdot t \cdot \cos 36^(\circ)`

`9.10\pm 0.23 = 0.809\cdot v_(o)\cdot t` (Ec. 1)

`2.70\,m = 2.20\,m + v_(o)\cdot t \cdot \sin 36^(\circ) + (1)/(2)\cdot \left(-9.807\,(m)/(s^(2)) \right) \cdot t^(2)`

`0.50 = 0.588\cdot v_(o)\cdot t-4.904\cdot t^(2)` (Ec. 2)

At first we clear
`v_(o)\cdot t` in (Ec. 1):

`v_(o)\cdot t = (9.10\pm 0.23)/(0.809)`

`v_(o)\cdot t = 11.248\pm 0.284`

(Ec. 1) in (Ec. 2):

`0.5 = 0.588\cdot (11.248\pm 0.284)-4.904\cdot t^(2)`

Now we clear the time in the resulting expression:

`4.904\cdot t^(2) = 0.588\cdot (11.248\pm 0.284)-0.5`

`t = \sqrt{(0.588\cdot (11.248\pm 0.284)-0.5)/(4.904) }`

There are two solutions:

`t_(1) = \sqrt{(0.588\cdot (11.248- 0.284)-0.5)/(4.904) }`

`t_(1) \approx 1.101\,s`

`t_(2) = \sqrt{(0.588\cdot (11.248+ 0.284)-0.5)/(4.904) }`

`t_(2)\approx 1.131\,s`

The initial velocity is cleared within (Ec. 2):

`v_(o)=(0.50+4.904\cdot t^(2))/(0.588\cdot t)`

The bounds of the range of initial speed is determined hereafter:

`t_(1) \approx 1.101\,s`

`v_(o) = (0.50+4.904\cdot (1.101)^(2))/(0.588\cdot (1.101))`

`v_(o) = 9.954\,(m)/(s)`

`t_(2)\approx 1.131\,s`

`v_(o) = (0.50+4.904\cdot (1.131)^(2))/(0.588\cdot (1.131))`

`v_(o) = 10.185\,(m)/(s)`

The range of initial speeds allowed to make the basket is:
`9.954\,(m)/(s)\leq v \leq 10.185\,(m)/(s)`.