New attempt is in progress. Some of the new entries may impact the last attempt grading. Incorrect. The hammer throw is a track-and-field event in which a 7.30-kg ball (the hammer) is whirled around in a circle several times and released. It then moves upward on the familiar curved path of projectile motion and eventually returns to the ground some distance away. The world record for the horizontal distance is 86.75 m, achieved in 1986 by Yuriy Sedykh. Ignore air resistance and the fact that the ball was released above the ground rather than at ground level. Furthermore, assume that the ball is whirled around a circle that has a radius of 2.22 m and that its velocity at the instant of release is directed 47.9° above the horizontal. Find the magnitude of the centripetal force acting on the ball just prior to the moment of release.

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hamzaahmeds hamzaahmeds · Answer 1 · 2021-10-16T08:28:32+02:00

This question involves the concepts of centripetal force, range of projectile and projectile motion.

The magnitude of centripetal force is "2812.8 N".

First, we will find the velocity of the ball by using the formula of the range of the projectile.

[tex]R = \frac{v^2Sin2\theta}{g}[/tex]

where,

R = range of projectile = 86.75 m

v = speed = ?

θ = launch angle = 47.9°

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]86.75\ m = \frac{(v)^2Sin(2)(47.9^o)}{9.81\ m/s^2}\\\\v = \sqrt{\frac{(86.75\ m)(9.81\ m/s^2)}{Sin95.8^o}}[/tex]

v = 29.25 m/s

Now, we will use the formula to find out the centripetal force:

[tex]F_c = \frac{mv^2}{r}[/tex]

where,

[tex]F_c[/tex] = Centripetal Force = ?

m = mass of the ball = 7.3 kg

v = speed = 29.25 m/s

r = radius = 2.22 m

Therefore,

[tex]F_c = \frac{(7.3\ kg)(29.25\ m/s)^2}{2.22\ m}[/tex]

Fc = 2812.8 N = 2.812 KN

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