Mighty Mouse (who flies at a constant speed of 600mph) was standing by a cannon that was pointed straight up. A cannonball was shot vertically into the air with an initial velocity of 300 mph. Exactly one second after the cannon was fired, Mighty Mouse flew up to touch the bottom of the cannonball. He then immediately turned and flew back down to touch the inside bottom of he cannon, then up to the cannonball, down into the cannon, and son on, until the cannonball fell back into the cannon's barrel.
What was the total distance Mighty Mouse ha flown when he was finally smashed into mouse-mush by the cannonball?
the height of the ball is
h(t) = 134t - 9.8t^2
since 300 mph = 134m/s
so, when is the cannon ball back down? when h=0
t = 13.67 sec
so, how far does Mickey fly in that time (less 1 second), at 600 mph?
oops. that should be
h(t) = 134t - 4.9t^2
so solve that for the correct time involved.
To find the total distance Mighty Mouse flew before he was smashed by the cannonball, we need to determine the distance he traveled in each round trip and then add them up.
Let's break down the problem step by step:
Step 1: Find the time it takes for the cannonball to fall back into the cannon's barrel.
Since the cannonball was shot vertically with an initial velocity of 300 mph, we can find the time it takes to reach its peak using the formula:
\[t = \frac{v_f - v_i}{a}\]
where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, and \(a\) is the acceleration. In this case, the cannonball is moving vertically, so \(a = -32 \, \text{ft/s}^2\) (assuming Earth's gravity).
Converting the velocities to feet per second:
\(v_f = \frac{300 \, \text{mph} \times 5280 \, \text{feet}}{3600 \, \text{s}} = 440 \, \text{ft/s}\)
\(v_i = 0 \, \text{ft/s}\)
\(t = \frac{440 \, \text{ft/s} - 0 \, \text{ft/s}}{-32 \, \text{ft/s}^2} = -13.75 \, \text{s}\)
Since time cannot be negative, we take the absolute value of \(t\), giving us 13.75 seconds.
Step 2: Find the distance Mighty Mouse travels in one round trip.
To find the distance Mighty Mouse travels in one round trip, we need to calculate the sum of the distances he traveled up and down to touch the bottom of the cannonball and the bottom of the cannon.
Since Mighty Mouse and the cannonball are both flying at a constant speed, their distances can be calculated using the formula:
\[d = v \times t\]
where \(d\) is the distance, \(v\) is the velocity, and \(t\) is the time.
In this case, Mighty Mouse is flying at a constant speed of 600 mph, which is equal to 880 ft/s (converting from mph to ft/s).
Since Mighty Mouse and the cannonball meet at the same position, we can split the time evenly between the upward and downward journeys. Therefore, Mighty Mouse spends half of the total time (6.875 seconds) to reach the highest point and half to reach the cannonball again.
For the upward journey, Mighty Mouse travels:
\(d_1 = v \times t_1 = 880 \, \text{ft/s} \times 6.875 \, \text{s} = 6050 \, \text{ft}\)
For the downward journey, Mighty Mouse travels:
\(d_2 = v \times t_2 = 880 \, \text{ft/s} \times 6.875 \, \text{s} = 6050 \, \text{ft}\)
Therefore, the total distance Mighty Mouse travels in one round trip is:
\(d_\text{total} = d_1 + d_2 = 6050 \, \text{ft} + 6050 \, \text{ft} = 12100 \, \text{ft}\)
Step 3: Calculate the total distance Mighty Mouse flies.
Since Mighty Mouse keeps making round trips as long as the cannonball is in the air, we can find the total distance by multiplying the number of round trips (which is the same as the number of seconds) by the distance per round trip.
Since the cannonball takes 13.75 seconds to fall back into the cannon's barrel, the total distance Mighty Mouse flies is:
\(d_\text{total} = 13.75 \, \text{s} \times 12100 \, \text{ft/s} \approx 166,375 \, \text{ft}\)
Therefore, Mighty Mouse flies approximately 166,375 feet before being smashed by the cannonball.