Tom Brady (the Patriots’ quarterback) throws a football – at a speed of 30 mph at an angle of 50 degrees to the horizontal – toward an intended receiver 20 yd straight downfield behind the goal-line. The pass is released 7.5 ft above the ground. Assume that the receiver is stationary, is isolated and that he can catch the ball from the ground (a shoestring catch) up to a height of 7 ft when the ball arrives at his position. Will the pass be completed? If not will it be short or long? If the conditions (launch height, distance to receiver, height for catch) are exactly the same but the quarterback can only launch the football at 60 degrees to the horizontal (over the outstretched arms of a left guard) what is the range of speeds (accurate to 3 significant digits) he must give the ball at launch so as to make a completion?
To determine if the pass will be completed, we need to analyze the motion of the football and see if it reaches a height within the catchable range of the receiver.
First, we need to convert the given measurements to the appropriate units.
30 mph = (30 * 5280 ft) / (3600 s) = 44 ft/s
20 yd = 60 ft
7.5 ft = 7.5 ft
7 ft = 7 ft
Now, let's calculate the time it takes for the football to reach the receiver's position. We can use the following equation:
d = v₀ * t + 0.5 * a * t²
Where:
d = distance traveled (60 ft)
v₀ = initial velocity (44 ft/s)
a = acceleration (assume g = 32.2 ft/s², assuming no air resistance)
t = time
Rearranging the equation, we have:
t = (sqrt(2 * d / a))
t = (sqrt(2 * 60 ft / 32.2 ft/s²))
t ≈ 1.05 s
Now, let's calculate the maximum height the football reaches using the following equation:
h = v₀ * sin(θ) * t - 0.5 * g * t²
Where:
h = maximum height
v₀ = initial velocity (44 ft/s)
θ = launch angle (50 degrees)
t = time
h = (44 ft/s) * sin(50 degrees) * (1.05 s) - 0.5 * (32.2 ft/s²) * (1.05 s)²
h ≈ 33.47 ft
From the calculation, we can see that the maximum height the football reaches is 33.47 ft. As this is higher than the receiver's catchable range of 7 ft, the pass will be completed.
Now, let's consider the scenario where the launch angle is 60 degrees. We need to find the range of speeds for the pass to be completed.
Using the same equation for time, we can calculate the time it takes for the football to reach the receiver's position:
t = (sqrt(2 * 60 ft / 32.2 ft/s²))
t ≈ 1.05 s
To find the range of speeds, we need to determine the initial velocity that will result in a maximum height within the receiver's catchable range of 7 ft.
Using the equation for maximum height, we have:
h = (v₀ * sin(θ) * t) - 0.5 * (32.2 ft/s²) * t²
Rearranging the equation, we get:
v₀ * sin(θ) = (h + 0.5 * (32.2 ft/s²) * t²) / t
v₀ = ((h + 0.5 * (32.2 ft/s²) * t²) / t) / sin(θ)
v₀ = ((7 ft + 0.5 * (32.2 ft/s²) * (1.05 s)²) / (1.05 s)) / sin(60 degrees)
v₀ ≈ 61.97 ft/s
Therefore, to make a completion with a launch angle of 60 degrees, the quarterback must give the ball a speed of approximately 61.97 ft/s.
To determine whether the pass will be completed, we need to analyze the projectile motion of the football.
First, let's convert the given units:
- 30 mph = 44 ft/s
- 20 yd = 60 ft
- 7.5 ft = 2.286 m (launch height)
- 7 ft = 2.134 m (receiver's catch height)
Now, let's break down the problem into components:
1. Horizontal Motion:
The horizontal motion of the football does not have any acceleration, as there is no force acting on it in that direction. Thus, the horizontal velocity remains constant throughout the motion. We can calculate the initial horizontal velocity (Vx) using the given angle and speed:
Vx = Velocity * cos(angle)
Vx = 44 ft/s * cos(50 degrees) = 28.26 ft/s
2. Vertical Motion:
The vertical motion is affected by the acceleration due to gravity (-9.8 m/s²). We need to calculate the time it takes for the football to reach the receiver. We can use the equation of motion:
Δy = Vyi * t + (1/2) * a * t²
where Δy is the change in vertical position, Vyi is the initial vertical velocity, t is the time, and a is the acceleration.
The change in vertical position is the difference between the launch height and the receiver's catch height:
Δy = 2.286 m - 2.134 m = 0.152 m
Plugging in the values, we get:
0.152 m = (Vyi * t) + (0.5) * (-9.8 m/s²) * t²
0.152 m = Vyi * t - 4.9 m/s² * t²
Since we don't know the initial vertical velocity (Vyi) and time (t) separately, we can eliminate one of the variables by using trigonometry.
3. Vertical and Horizontal Motions:
The time taken (t) can be calculated using the equation:
t = Δx / Vx
where Δx is the horizontal distance.
Plugging in the values, we get:
t = 60 ft / 28.26 ft/s = 2.121 s
Now, we can substitute this value of t back into the earlier equation to solve for Vyi:
0.152 m = Vyi * (2.121 s) - 4.9 m/s² * (2.121 s)²
Simplifying the equation, we get:
0.152 m = Vyi * 2.121 s - 20.739 m
To complete the pass, the receiver should be able to catch the ball when it is at a height of 2.134 m. However, the equation tells us that the ball will land at a height of 0.152 m.
Therefore, the pass will not be completed, and it will be short of the receiver.
Moving on to the second part of the question:
If the quarterback launches the football at a 60-degree angle, we need to find the range of speeds (accurate to 3 significant digits) for a completion.
To find the range of launch speeds, we need to analyze the vertical motion. The time taken (t) can be calculated using the same horizontal distance (Δx) as before:
t = Δx / Vx
Now, we can substitute this time (t) back into the equation for vertical motion to solve for Vyi:
Δy = Vyi * t - 4.9 m/s² * t²
Substituting the launch and catch heights:
0.152 m = Vyi * t - 4.9 m/s² * t²
2.134 m = Vyi * t - 4.9 m/s² * t²
Here, we have two equations with two unknowns: Vyi and t. We can solve them simultaneously to find the launch speed.
Using the first equation:
t = Δx / Vx
t = 60 ft / 28.26 ft/s = 2.121 s
Substituting this value back into the second equation:
2.134 m = Vyi * (2.121 s) - 4.9 m/s² * (2.121 s)²
Simplifying the equation, we get:
2.134 m = Vyi * 2.121 s - 20.739 m
Solving for Vyi, we have:
Vyi * 2.121 s = 20.739 m + 2.134 m
Vyi = (20.739 m + 2.134 m) / 2.121 s
Vyi = 11.03 m/s (approximately)
Therefore, for a completion with a launch angle of 60 degrees, the quarterback must give the ball a speed of approximately 11.03 m/s (accurate to 3 significant digits).