What Is the Radial Acceleration in an Olympic Hammer Throw?

AI Thread Summary
In the Olympic hammer throw, the radial acceleration just before release is calculated using the mass of the hammer, the length of the wire, and the height from which it is released. The problem states that the hammer is released from a height of 1.3 m and travels 84 m horizontally at an angle of 24 degrees. The initial calculations yield a radial acceleration of 955 m/s², but the book states the correct answer is 892 m/s². The discrepancy suggests a need for reevaluation of the velocity and height equations used. Accurate calculations are essential for determining the correct radial acceleration in this projectile motion scenario.
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[SOLVED] circular and projectile motion

Homework Statement


In the Olympic hammer throw, contestants whirl a 7.3 kg ball on the end of a 1.2 m long wire before releasing it. Suppose the hammer is released frim a height of 1.3 m while moving in a direction 24 degrees above the horizontal. If it travels 84 m horizontally, what is its radial acceleration just before release.
The book answer is a = 892 m/s^2


Homework Equations


y = x tan theta - gx^2/2v0^2 cos^2 theta

a = v^2/r

The Attempt at a Solution



1.3 = 84 (.445) - 9.8 (84)^2/2v0^2 (.914)^2

v0^2 = 41386.8/36.1 = 1146.4

a = 1146.4/1.2 = 955 m/s^2

Even if I just use the velocity in the x direction, it does not work.
 
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The starting height is 1.3 m. The height when x=84 is 0 m
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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