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nirvana33 [79]
3 years ago
13

A place kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. half the crowd hopes the ball will clear

the crossbar, which is 3.05 m high. when kicked, the ball leaves the ground with a speed of 23.4 m/s at an angle of 50.0° to the horizontal. (a) by how much does the ball clear or fall short of clearing the crossbar
Physics
1 answer:
trapecia [35]3 years ago
4 0
<span>The ball clears by 11.79 meters Let's first determine the horizontal and vertical velocities of the ball. h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s Now determine how many seconds it will take for the ball to get to the goal. t = 36.0 m / 15.04 m/s = 2.394 s The height the ball will be at time T is h = vT - 1/2 A T^2 where h = height of ball v = initial vertical velocity T = time A = acceleration due to gravity So plugging into the formula the known values h = vT - 1/2 A T^2 h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2 h = 42.92 m - 4.9 m/s^2 * 5.731 s^2 h = 42.92 m - 28.0819 m h = 14.84 m Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
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A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res
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Answer:

The time constant is \tau = 17.5 \ s    

Explanation:

From the question we are told that

   The spring constant is  k = 11.5 \  N/m

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   The amplitude of the  oscillation t the beginning is x =  6.70 cm = 0.067 \  m

    The amplitude after time t is  x_t = 2.20 cm = 0.022 \  m

    The number of oscillation is N  = 30

Generally the time taken to attain the second amplitude is mathematically represented as

       t  = N  *  T                                            Here  T is the period of oscillation

         t = N * 2\pi \sqrt{\frac{m}{k} }

=>     t = 30 * 2 * 3.142 *  \sqrt{\frac{ 0.490}{11.5} }

=>     t = 38.88 \  s

Generally the amplitude at time t is mathematically represented as

         x(t) = x e^{-\frac{at}{2m} }

Here a is the damping  constant so

 at  t = 38.88 \  s ,  x_t = 2.20 cm = 0.022 \  m

So  

     0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }

=>  0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }

taking natural log of both sides

=>  ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }    

=>   a = 0.028

Generally the time constant is mathematically represented as

    \tau = \frac{m}{a}      

=> \tau = \frac{0.490}{  0.028}    

=> \tau = 17.5 \ s    

4 0
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Answer:

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Explanation:

Answer is above

<em><u>Hope this helps.</u></em>

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