Rotational Kinematics Football Problem

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SUMMARY

The problem involves calculating the number of revolutions a football makes while in the air after being thrown by a quarterback. The ball spins at 9.0 revolutions per second (rev/s) and is thrown with a linear speed of 21 m/s at an angle of 51° from the ground. The correct calculation shows that the ball completes 30 revolutions during its flight, which is derived from considering the total time of flight rather than just the ascent. The initial attempt incorrectly calculated only the ascent time, resulting in an answer of 15 revolutions.

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Homework Statement



A quarterback throws a pass that is a perfect spiral. In other words, the football does not wobble, but spins smoothly about an axis passing through each end of the ball. Suppose the ball spins at 9.0 rev/s. In addition, the ball is thrown with a linear speed of 21 m/s at an angle of 51° with respect to the ground. If the ball is caught at the same height at which it left the quarterback's hand, how many revolutions has the ball made while in the air?

Vf = 21 m/s
Vo = 0 m/s
θ = 51°
ω = 9.0 rev/s
a = 9.81m/s^2

Homework Equations



ω = Δθ/Δt

V = Vo + at

The Attempt at a Solution



I derived this from the kinematic equation
V-Vo/a = t

so then I have
(sin(51°)*21/ms))/9.81m/s^2 = 1.66

then
9.0*1.66 = 15 rev

The problem is that answer is wrong and I have NO idea where I went wrong...any help would be greatly appreciated.

The actual answer is 30rev
 
Last edited:
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lolcheelol said:

Homework Statement



A quarterback throws a pass that is a perfect spiral. In other words, the football does not wobble, but spins smoothly about an axis passing through each end of the ball. Suppose the ball spins at 9.0 rev/s. In addition, the ball is thrown with a linear speed of 21 m/s at an angle of 51° with respect to the ground. If the ball is caught at the same height at which it left the quarterback's hand, how many revolutions has the ball made while in the air?

Vf = 21 m/s
Vo = 0 m/s
θ = 51°
ω = 9.0 rev/s
a = 9.81m/s^2

Homework Equations



ω = Δθ/Δt

V = Vo + at

The Attempt at a Solution



I derived this from the kinematic equation
V-Vo/a = t

so then I have
(sin(51°)*21/ms))/9.81m/s^2 = 1.66

then
9.0*1.66 = 15 rev

The problem is that answer is wrong and I have NO idea where I went wrong...any help would be greatly appreciated.

The actual answer is 30rev

If you have half the answer, perhaps you only considered the time taken for the ball to reach maximum height, and forgot about it coming back down again ?
 

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