In 1939 or 1940, Emanuel Zacchini took his human-cannonball act to an extreme: After being shot from a cannon, he soared over three Ferris wheels and into a net (see the figure). Assume that he is launched with a speed of 28 m/s and at an angle of 56°. (a) Treating him as a particle, calculate his clearance over the first wheel. (b) If he reached maximum height over the middle wheel, by how much did he clear it? (c) How far from the cannon should the net's center have been positioned (neglect air drag)?
To solve this problem, we need to break it down into three parts:
(a) Calculating his clearance over the first wheel:
To calculate his clearance over the first wheel, we need to determine the maximum height he reaches. We can use the kinematic equation for vertical motion:
v^2 = u^2 - 2as
Where:
v = final vertical velocity (0 m/s at maximum height)
u = initial vertical velocity (vertical component of launch velocity)
a = acceleration (acceleration due to gravity, -9.8 m/s^2)
s = vertical displacement (maximum height reached)
We can find the initial vertical velocity by multiplying the launch speed (28 m/s) by the sine of the launch angle (56°).
u = 28 m/s * sin(56°)
Plugging in the values into the kinematic equation:
0^2 = (28 m/s * sin(56°))^2 - 2 * (-9.8 m/s^2) * s
Simplifying the equation:
0 = (28^2 m^2/s^2 * sin^2(56°)) + 19.6 m/s^2 * s
Now we can solve for s:
s = - (28^2 m^2/s^2 * sin^2(56°)) / (19.6 m/s^2)
Using a calculator, we find that s ≈ 20.79 m.
Therefore, his clearance over the first wheel is approximately 20.79 meters.
(b) Finding the clearance over the middle wheel:
To find the clearance over the middle wheel, we need to determine the maximum height he reaches. Using the same equation as above:
0 = (28 m/s * sin(56°))^2 - 2 * (-9.8 m/s^2) * s
However, this time we want to find the maximum height, so we need to solve for u:
u = √((28 m/s * sin(56°))^2 + 2 * 9.8 m/s^2 * s)
Using the same initial launch velocity as before (28 m/s * sin(56°)), we can find the maximum height.
Substituting the known values into the equation:
u = √((28 m/s * sin(56°))^2 + 2 * 9.8 m/s^2 * s)
Plugging in the values and using a calculator, we find that u ≈ 31.68 m/s.
Therefore, he reaches a maximum height of approximately 31.68 meters over the middle wheel.
To find the clearance, we need to subtract the radius of the middle wheel from the maximum height:
Clearance = 31.68 m - radius of the middle wheel.
(c) Calculating the position of the net's center:
To find the distance from the cannon to the net's center, we can use the horizontal motion of the cannonball. Since there is no horizontal acceleration, the horizontal velocity remains constant throughout the motion.
v = u * cos(θ)
Where:
v = horizontal velocity
u = initial launch speed (28 m/s)
θ = launch angle (56°)
Plugging in the values:
v = 28 m/s * cos(56°)
Using a calculator, we find that v ≈ 15.15 m/s.
Next, we can use the equation for horizontal distance:
distance = velocity * time
Assuming the time taken to reach the net is the same as the time taken to reach the maximum height, we can use the formula:
distance = 15.15 m/s * (total time / 2)
To find the total time, we first calculate the time to reach maximum height using the equation:
v = u + at
Where:
v = final vertical velocity (0 m/s at maximum height)
u = initial vertical velocity (vertical component of launch velocity)
a = acceleration (acceleration due to gravity, -9.8 m/s^2)
t = time
Plugging in the values:
0 = (28 m/s * sin(56°)) + (-9.8 m/s^2) * t
Simplifying the equation:
t = (28 m/s * sin(56°)) / (9.8 m/s^2)
Using a calculator, we find that t ≈ 3.37 s.
Therefore, the total time is approximately 3.37 s * 2 = 6.74 s.
Plugging in the time into the distance equation:
distance = 15.15 m/s * 6.74 s / 2
Using a calculator, we find that the distance is approximately 51.07 meters.
Therefore, the net's center should have been positioned approximately 51.07 meters away from the cannon.
Please note that these calculations neglect air drag, which may affect the actual values in a real-world scenario.