Mighty Mouse (who flies at a constant speed of 1400 ft/sec) was standing next to a cannon that was pointed straight up. A cannonball was shot vertically into the air with an initial velocity of 880 ft/sec. Exactly one second after the cannon was fired, Might Mouse flew up to touch the bottom of the cannonball. He then immediately turned and flew back down to touch the inside bottom of the cannon, then up to the cannonball, down into the cannon, and so on, until the cannonball fell back in the cannon's barrel. What was the total distance Mighty Mouse had flown when he was finally smashed into mouse-mush by the cannonball?
the cannonball: v=880-32t
v=0 (max height) at t=27.5 sec
total time in air: 55 sec
so, Mickey flew for 54 sec at 1400ft/sec = 75600 ft
assuming zero body thickness and reach for Mickey.
Otherwise he could not fly during the final instants when the ball was crushing his body, and his flights between cannon bottom and ball were shorter than the distance between them.
sorry. Mighty. Mickey does not fly. :-(
To determine the total distance Mighty Mouse flew, we need to calculate the distance he covered during each cycle until the cannonball fell back into the barrel.
First, let's consider the motion of the cannonball. Since it was shot vertically with an initial velocity of 880 ft/sec and experienced only the force of gravity, we can use the equation of motion:
y = y0 + v0t - 0.5gt^2
where:
y = final position (height of the cannonball)
y0 = initial position (0 ft)
v0 = initial velocity (880 ft/sec)
t = time (seconds)
g = acceleration due to gravity (32 ft/sec^2)
We can re-arrange the equation to find the time it takes for the cannonball to fall back into the barrel:
0 = 0 + 880t - 0.5(32)(t^2)
16t^2 - 880t = 0
t(16t - 880) = 0
Solving for t, we have two possible solutions: t = 0 (initial moment) and t = 55 seconds (when the cannonball falls back into the barrel). However, in this case, we only consider t = 55 seconds because it represents the full cycle.
During the one-second delay before Mighty Mouse started flying, the cannonball traveled for 55 - 1 = 54 seconds.
Now, let's calculate the distance Mighty Mouse flew during each cycle:
1. From the initial moment (t = 1 second) until the cannonball meets him:
Mighty Mouse flies at a constant speed of 1400 ft/sec, so the distance he covers is:
(1400 ft/sec) * (t - 1 second) = 1400(t - 1) ft
2. From the moment the cannonball meets Mighty Mouse until he touches the inside bottom of the cannon:
The cannonball continues to move upwards and meets Mighty Mouse at t = 55 seconds. During this time, Mighty Mouse's upward speed equals the cannonball's downward speed (880 ft/sec). Therefore, the time taken for this motion is t = (55 - 1) / 2 = 27 seconds.
The distance Mighty Mouse covers is:
(880 ft/sec) * (27 seconds) = 23760 ft
3. From the moment Mighty Mouse touches the inside bottom of the cannon until he reaches the cannonball:
During this time, Mighty Mouse descends at a constant speed of 1400 ft/sec. The time taken for this motion is also 27 seconds, as it corresponds to the time it took to reach the cannonball.
The distance Mighty Mouse covers is:
(1400 ft/sec) * (27 seconds) = 37800 ft
Adding up the distances covered in each phase, the total distance Mighty Mouse flew is:
1400(t - 1) + 23760 + 37800 = 1400t + 61560 ft
Substituting t = 55 seconds:
1400(55) + 61560 = 77000 + 61560 = 138560 ft
Therefore, Mighty Mouse flew a total distance of 138,560 feet before being smashed into mouse-mush by the cannonball.