Question

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 flat 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?

Answer 1

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      

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

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

=>        $F = 168.298 \ N$