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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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