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3 years ago

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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?

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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Let's solve the problem in the three different steps

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