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mojhsa [17]
2 years ago
9

To win the game, a place kicker must kick a football from 30m from the goal, and the ball must clear the crossbar which is 3.05

m high. When kicked, the ball leaves the ground with the speed of 19 m/s at the angle of 37.2° from the horizontal. Acceleration of gravity is 9.8 m/s^2
How much vertical distance does the ball clear the crossbar?
Physics
1 answer:
Mashcka [7]2 years ago
7 0

Answer:

Explanation:

Ignore air resistance

The horizontal velocity is

vx =19cos37.2 = 15.134 m/s

The ball travels 30 m in a time of

30 / 15.134 = 1.982 s

In that time, the ball will be at an altitude of

s = vy₀t + ½at²

s = 19sin37.2(1.982) + ½(-9.8)1.982² = 3.5137... m

so the ball clears the bar by

h = 3.5137 - 3.05 = 0.4637... = 0.46 m

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What force must the deltoid muscle provide to keep the arm in this position?
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Answer:

Deltoid Force, F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

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

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       ⇒           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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Explanation:

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