Rotational Kinematics Football Problem

AI Thread Summary
The problem involves a quarterback throwing a football that spins at 9.0 revolutions per second and travels at a linear speed of 21 m/s at a 51° angle. The initial attempt to calculate the number of revolutions made while in the air resulted in 15 revolutions, which was incorrect. The correct answer is 30 revolutions, suggesting that the initial calculation may have only accounted for the time to reach maximum height without considering the descent. This highlights the importance of including the entire flight duration in kinematic calculations.
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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
 
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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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