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3 years ago

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The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing

on the free throw line throws the ball with an initial speed of 7.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved in solving projectile motion problems.

1 answer:

Mars2501 [29]3 years ago

7 0

Answer:

49.2°

Explanation:

Using the equation of the trajectory

y = y₀ + (x - x₀) tanθ - ( g(x-x₀)²) / 2( v₀cosθ)²

1 / cos θ = sec θ

y - y₀ = (x - x₀) tanθ - ( g(x-x₀)²) sec²θ / 2v²

y - y₀ = 3.05 m - 2.44 m ( height of the basket - height the thrower released the ball) = 0.61 m

x - x₀ = 4.57

g = 9.81 m/s²

v = 7.15 m/s

0.61 m = 4.57 tanθ - (9.81 × 4.57²) sec²θ / 2 ×7.15²

0.61 m = 4.57 tanθ - 2 sec²θ

but sec²θ = tan²θ + 1

0.61 m = 4.57 tanθ - 2 (tan²θ + 1)

0.61 m = 4.57 tanθ - 2tan²θ - 2

0 = - 2tan²θ + 4.57 tanθ - 2 - 0.61 m

0 = - 2tan²θ + 4.57 tanθ - 2.61

multiply by -1 both side

0 = 2 tan²θ - 4.57 tanθ + 2.61

let x = tan θ

2 x² - 4.57 x + 2.61

using quadratic formula

-b ± √ ( b² - 4ac) / 2a

(- (- 4.57) ± √( (-4.57)² - (4 × 2 × 2.61))) / 2× 2

1.16 or 1.125

tanθ = 1.16

θ = tan⁻¹1.16 = 49.2°

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Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

$\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a$ (1)

Mass B

$\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a$ (2)

Where:

$T$ - Tension force in the cord, measured in newtons.

$m_{A}$ , $m_{B}$ - Masses of blocks A and B, measured in kilograms.

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

$a$ - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

$m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g$

$(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g$

$a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g$

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$a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)$

$a = 4.203\,\frac{m}{s^{2}}$

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

$T = m_{B}\cdot (a+g)$

If we know that $m_{B} = 2\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $a = 4.203\,\frac{m}{s^{2}}$ , then the tension force in the cord:

$T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)$

$T = 28.02\,N$

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

$\Delta y = \frac{1}{2}\cdot a\cdot t^{2}$ (3)

Where:

$a$ - Net acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

$\Delta y$ - Covered distance, measured in meters.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $\Delta y = 1.5\,m$ , then the time taken by the system is:

$t = \sqrt{\frac{2\cdot \Delta y}{a} }$

$t = \sqrt{\frac{2\cdot (1.5\,m)}{4.203\,\frac{m}{s^{2}} } }$

$t \approx 0.845\,s$

The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is calculated by the following formula:

$v = a\cdot t$ (4)

Where $v$ is the final speed of the system, measured in meters per second.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $t \approx 0.845\,s$ , then the final speed of the system is:

$v = \left(4.203\,\frac{m}{s^{2}} \right)\cdot (0.845\,s)$

$v = 3.551\,\frac{m}{s}$

The final speed of the system is 3.551 meters per second.

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