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 5.8 rev/s. In addition, the ball is thrown with a linear speed of 17 m/s at an angle of 53° 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?
To find out how many revolutions the ball has made while in the air, we need to determine the total time the ball is in the air and then calculate the number of revolutions based on the spin rate.
Step 1: Calculate the vertical component of the initial velocity.
The initial velocity of the ball can be decomposed into horizontal (Vx) and vertical (Vy) components. The vertical component is given by:
Vy = V0 * sin(θ)
Where V0 is the initial velocity magnitude (17 m/s) and θ is the angle of the ball's trajectory (53°).
Vy = 17 m/s * sin(53°)
Vy ≈ 13.09 m/s
Step 2: Calculate the time of flight.
The time the ball is in the air can be found using the vertical component of the velocity. The equation for the time of flight is:
2 * Vy / g
Where g is the acceleration due to gravity (approximately 9.8 m/s²).
Time of flight = 2 * 13.09 m/s / 9.8 m/s²
Time of flight ≈ 2.67 s
Step 3: Calculate the number of revolutions.
The number of revolutions is given by the product of the spin rate (in rev/s) and the time of flight (in seconds):
Number of revolutions = Spin rate * Time of flight
Number of revolutions = 5.8 rev/s * 2.67 s
Number of revolutions ≈ 15.486 rev
Therefore, the football makes approximately 15.486 revolutions while in the air.
To find the number of revolutions the ball has made while in the air, we need to calculate the time the ball is in the air and then multiply it by the spin rate.
First, let's find the time the ball is in the air. We can use the vertical motion of the ball to determine this. Since the ball is caught at the same height it left the quarterback's hand, we know that the vertical displacement of the ball is zero.
The equation for vertical displacement is given by:
Δy = V0y * t + (1/2) * g * t^2
Here, Δy is the vertical displacement, V0y is the initial vertical velocity, t is the time, and g is the acceleration due to gravity.
Since the initial vertical velocity is given by V0y = V0 * sinθ, where V0 is the initial speed and θ is the launch angle, we can rewrite the equation:
0 = (V0 * sinθ) * t + (1/2) * g * t^2
Now we have a quadratic equation in terms of t. We can solve this equation to find the time the ball is in the air. In this case, V0 = 17 m/s and θ = 53°.
Using the quadratic formula, the equation becomes:
(1/2) * g * t^2 + (V0 * sinθ) * t = 0
Simplifying further, we have:
(1/2) * (9.8 m/s^2) * t^2 + (17 m/s * sin(53°)) * t = 0
Now we can solve for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = (1/2) * (9.8 m/s^2), b = 17 m/s * sin(53°), and c = 0.
Plugging in the values, we get:
t = (-17 m/s * sin(53°) ± √((17 m/s * sin(53°))^2 - 4 * (1/2) * (9.8 m/s^2) * 0)) / (2 * (1/2) * (9.8 m/s^2))
Calculating this equation gives two solutions, t = 1.66 s and t = 0 s. Since we want the time the ball is in the air, we discard the solution t = 0 s.
Therefore, the ball is in the air for approximately t = 1.66 seconds.
Now we can calculate the number of revolutions the ball has made using the spin rate of 5.8 rev/s. Multiply the spin rate by the time:
Number of revolutions = spin rate * time
Number of revolutions = 5.8 rev/s * 1.66 s
Number of revolutions ≈ 9.628 rev
Therefore, the ball has made approximately 9.628 revolutions while in the air.
The time in the air is 2*Vsin53/g
Multiply that by 5.8 rev/s for the number of revolutions.