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LUCKY_DIMON [66]
1 year ago
12

Rick shoots a basketball at an angle of 35' from the horizontal. It leaves his hands 7 feet from the ground with a velocity of 2

0 ft/s.Step 1 of 2: Construct a set of parametric equations describing the shot. Round all final values to the nearest tenth.Answer

Physics
1 answer:
Korvikt [17]1 year ago
8 0

Given:

The angle of projection of the basketball, θ=35°

The height at which the ball leaves the hand, h=7 ft

The initial velocity of the basketball, v=20 ft/s

To find:

The parametric equations describing the shot.

Explanation:

The range, x of the basketball is given by,

x=v\cos\theta t

On substituting the known values,

\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}

The change in the height, y of the basketball is given by,

y=-v\sin\theta t+\frac{1}{2}gt^2

Where g is the acceleration due to gravity.

On substituting the known values,

\begin{gathered} y=-20\times\sin35\degree\times t+\frac{1}{2}\times32\times t^2 \\ \implies y=-11.5t+16t^2 \end{gathered}

Final answer:

The parametric equations describing the shot are

\begin{gathered} \begin{equation*} x=16.4t \end{equation*} \\ \begin{equation*} y=-11.5t+16t^2 \end{equation*} \end{gathered}

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What force must the deltoid muscle provide to keep the arm in this position?
ruslelena [56]

Answer:

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Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.  

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, T_{Deltoid}

2. The torque of the arm, T_{arm}  

Assuming the arm is just being stretched and there is no rotation going on,

                        T_{Deltoid} = 0

                       T_{arm} = 0

       ⇒           T_{Deltoid} = T_{arm}

                  r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}

Where,

r_{d} is radius of the deltoid

F_{d} is the force of the deltiod

\alpha_{d} is the angle of the deltiod

r_{a} is the radius of the arm

F_{a} is the force of the arm , F_{a} = mg  which is the mass of the arm and acceleration due to gravity

\alpha_{a} is the angle of the arm

The force of the deltoid muscle is,

                                 F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}

but F_{a} = mg ,

                ∴            F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

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