Question

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 6.3 rev/s. In addition, the ball is thrown with a linear speed of 29 m/s at an angle of 45 ° 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?

Answer 1

Given that :

initial speed of the ball, v₀ = 29 m/s

making an angle with respect to the ground, = 45 degree

on vertical plane, initial vertical velocity i given by :

v_o,y = v₀ sin                                                                  { eq.1 }

inserting the values in eq.1,

v_o,y = (29 m/s) sin 45⁰

v_o,y = 20.5 m/s

Using equation of motion 1, we have

v = v_o,y + g t                                                 { eq.2 }

where, g = acceleration due to gravity downward = -9.8 m/s²

v = final velocity of the ball = 0 m/s

inserting the values in eq.2,

(0 m/s) = (20.5 m/s) - (9.8 m/s²) t

t = (20.5 m/s) / (9.8 m/s²)

t = 2.09 sec

Total time taken by ball, t = 2 (2.09 sec) 4.18 sec

Total number of revolutions by the ball while in the air which will be given as :

Suppose, the ball spins at 6.3 rev/s.

N = (6.3 rev/s) (4.18 s)

N = 26.3 rev