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Rufina [12.5K]
3 years ago
12

A football is kicked straight up in the air , it hits the ground 5.6 s later. What was the greatest height kicked reached by the

ball? Assume it was kicked at ground level.
Physics
2 answers:
marin [14]3 years ago
5 0

Answer: 38.5 m

Explanation:

1) Flight time = 5.6 s (given)

2) Ascending time = fligth time / 2 = 5.6s / 2 = 2.8 s

3) Vf = Vo - gt

Hmax ⇒ Vf = 0

⇒ 0 = Vo - gt ⇒ Vo = gt = 9.81m/s² (2.8s) = 27.5 m/s

4) H = Vot - gt² / 2

⇒ H = 27.5m/s(2.8s) - 9.81m/s² (2.8s)² / 2 = 77m - 38.5m = 38.5m

Answer: 38.5m

Reil [10]3 years ago
3 0
We'll just consider the down trip, from when it was at its highest point to when it hits the ground.  Initial velocity is 0 m/s, acceleration is 9.81 m/s^2 (gravity), and time is 5.6s.

One of the kinematic equations is d = (Vi)(t) + 0.5(a)(t^2).
Plugging it in gives us d = 0 + (0.5)(9.81)(5.6^2) = 153.8208 m.

Hope this helped!
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You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.
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Answer:

The distance is r_2  =  0.24 \  m

Explanation:

From the question we are told that

       The  distance from the conversation is r_1    =  24.0 \ m

       The  intensity of  the sound at your position is  \beta _1 =  40 dB

        The  intensity at the sound at the new position is  \beta_2 =  80.0dB

Generally the intensity in  decibel is  is mathematically represented as

      \beta  =  10dB log_{10}[\frac{d}{d_o} ]

The intensity is  also mathematically represented as

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    \beta  =  10dB *  log_{10}[\frac{P}{A* d_o} ]

=>   \frac{\beta}{10}  =  log_{10} [\frac{P}{A (l_o)} ]

From the logarithm definition

=>    \frac{P}{A  *  d_o}  =  10^{\frac{\beta}{10} }

=>      P =  A (d_o ) [10^{\frac{\beta }{ 10} } ]

Here P is the power of the sound wave

 and  A is the cross-sectional area of the sound wave  which is generally in spherical form

Now the power of the sound wave at the first position is mathematically represented as

               P_1 =  A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]

Now the power of the sound wave at the second  position is mathematically represented as

               P_2 =  A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]

Generally  power of the wave is constant at both positions  so  

    A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]  = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]

      4 \pi r_1 ^2   [10^{\frac{\beta_1 }{ 10} } ]  = 4 \pi r_2 ^2   [10^{\frac{\beta_2 }{ 10} } ]

        r_2 =  \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}

       substituting value

        r_2 =   \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}

        r_2  =  0.24 \  m

     

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