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Viktor [21]
3 years ago
12

A golf ball (m=26.7g) is struck a blow that makes an angle of 33.6 degrees with the horizontal. The drive lands 190m away on a f

lat fairway. The acceleration of gravity is 9.8 m/s^2 . If the golf club and ball are in contact for 7.13 ms, what is the average force of impact?
Physics
1 answer:
koban [17]3 years ago
4 0

Answer:

Th average force impact is F = 168.298 \  N

Explanation:

From the question we are told that  

   The mass of the golf ball is  m_g =  26.7 \  g  =  0.0267 \  kg

    The angle made is \theta =  33.6 ^o

    The range of the golf ball  is R =  190 \  m

     The duration of contact is \Delta  t =  7.13 \  ms = 7.13 *10^{-3} \ s

Generally the range of the golf ball is mathematically represented as

       R = \frac{v^2 sin2(\theta)}{g}

Here v  is the velocity with which the golf club propelled it with, making  v the subject

       v  =  \sqrt{\frac{R *  g}{sin 2 (\theta)} }

=>     v  =  \sqrt{\frac{190 *  9.8}{sin 2 (33.6)} }

=>     v  =  44.94 \  m/s

Generally the change in momentum of the golf ball is mathematically represented as

      \Delta p  =  m  *  (v - u )

here u  is the initial  velocity of the ball before being stroked and the value is 0 m/s

       \Delta p  =  0.0267  *  ( 44.94 - 0 )

=>    \Delta p  = 1.19996 \  kg \cdot m/s

Generally the  average force of impact is mathematically represented as      

         F = \frac{\Delta  p }{\Delta t}

=>        F = \frac{1.19996 }{7.13 *10^{-3}}

=>        F = 168.298 \  N

     

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