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University of California, Berkeley Physics 8A, Section 1, Spring 2007 (Yury Kolomensky)

SOLUTIONS TO PROBLEM OF THE WEEK: # 1-#3

1. (20 points) Importance of Units: The Gimli $40M glider. Although the airplane had no power, the Throughout this course, we will expect you to be careful pilot managed to glide it toward an old airforce base in with units in your calculations. Yet some students tend Gimli, Manitoba. to neglect them and just trust that they always work out Under the circumstances, the glide and the landing properly. Maybe this real-world example will keep you were impeccable. Unfortunately, the runway at that air- from such a sloppy habit. The incident is known in avi- port had been converted to a track for race cars, and a ation as the Gimli Glider. steel barrier had been constructed across it. Fortunately, On July 23, 1983, Flight 143 was be- as the airplane hit the runway, the front col- ing readied for its long trip from to Edmon- lapsed, dropping the nose of the airplane onto the run- ton. The fuel gauges on board were not operational, and way. The skidding slowed the airplane so that it stopped the flight crew asked the ground crew to determine how just short of the steel barrier, with stunned race drivers much fuel was already on board. The flight crew knew and fans looking on. All on board the airplane emerged they needed to begin the trip with 22,300 kg of fuel. safely. The point here is this: Take care of the units. They knew that amount in kilograms because Canada 1. Solution had recently switched to the metric system; previously fuel had been measured in pounds. The ground crew (a) The crew thought they had could measure the onboard fuel only in liters, which they reported as 7,682 L. Thus, to determine how much 7, 682 L · 1.77 kg/L = 13, 600 kg fuel was on board and how much additional fuel was needed, the flight crew asked the ground crew for the of fuel on board. conversion factor from liters to kilograms of fuel. The (b) According to the crew’s calculations, they response was 1.77, which the flight crew used (assum- thought they needed only ing that 1.77 kg corresponds to 1 L). (a) How many kilograms of fuel did the flight crew 22, 300 − 13, 600 = 8, 700 kg think they had? (In this problem, take all given data as being exact.) of fuel, which, using the incorrect value for den- (b) How many liters did they ask to be added? sity, they thought would correspond to Unfortunately, the response from the ground crew 8, 700 kg = 4915 L was based on pre-metric habits: 1.77 was the conversion 1.77 kg/L factor not from liters to kilograms but rather from liters of jet fuel. to pounds of fuel (1.77 lb corresponds to 1 L). (c) How many kilograms of fuel were actually on board, (c) In reality, the aircraft had before the ground crew added some? (Except for the given 1.77, use four significant figures for other conver- 7, 682 L · 1.77 lb/L · 0.454 kg/lb = 6, 173 kg sion factors.) of fuel. (d) How many liters of additional fuel were actually needed? (d) Thus, they needed (e) When the airplane left Montreal, what percentage of (22, 300 − 6, 173) kg the required fuel did it have? = 20, 070 L 1.77 lb/L · 0.454 kg/lb On route to Edmonton, at an altitude of 41,000 feet, the airplane ran out of fuel and became a of fuel. 2

(e) When the jet left Montreal, the tanks were only (b) If you assume that the airplane travels the en- tire distance with the airspeed of 466 knots under 6, 173 + 4915 · 1.77 · 0.454 · 100% = 45.4% the headwind conditions above, how long would 22, 300 it take to fly between SFO and HNL ? Com- full, which meant the jet was doomed to run out pare this time to the flight time declared above: of fuel at about midpoint of the journey. this tells you how good the assumption is (and roughly how long it takes for takeoff and de- scent). 2. (30 points) Champagne from United 97 This is another real life story; one in which yours (c) Luckily for me, I noticed that we took off almost truly was able to gain an (unfair) advantage from his directly westward, and spent about 13 minutes physics problem-solving skills. to reach the cruising speed. What was the aver- United Airlines runs a little contest on every flight age acceleration and average ground speed dur- from the US mainland to Hawaii. The passengers are ing that time ? asked to determine the time the plane will reach the ex- act geographical midpoint of the journey (e.g. the point (d) Taking results from (c) into account and assum- exactly half distance between SFO and HNL airports). ing that the ground speed was constant after the The person who gets the closest answer wins a bottle of first 13 minutes, how long did it take to reach the good French champagne (typically reserved for those in midpoint of the journey ? First Class). I won the bottle on my recent (and the only) trip to Honolulu; I was off by 7 seconds (I do not mean Good luck ! to brag; as you may guess from the calculation below, 2. Solution there is an uncertainty of a few minutes in the answer, and hence some luck is involved). But let’s see if you (a) The pilot has given us the airspeed, i.e. the can reproduce the calculation. speed of the aircraft relative to moving air. Since Shortly after takeoff, the flight attendant announced the wind blows in the direction opposite to the the game, and gave the relevant (and some irrelevant) plane’s velocity (it is headwind), to determine data: the ground speed, you need to subtract the wind speed from the airspeed: • Total estimated flight time: 5 hrs 15 mins

v1st half = 466 − 33 = 433 knots • Departure (wheels up) time: 6:28pm PDT v2nd half = 466 − 40 = 426 knots • Flight distance SFO-HNL: 2128 nautical miles (1)

• Cruising airspeed: 466 knots (nautical miles per (b) If the entire journey were taken with the airspeed hour). This is the speed of the plane relative to above, the first half of the journey would take the air. 1064 mi t1st half = = 2.46 hours • Estimated headwind: 33 knots for the first half 433 mph of the journey, 40 knots for the second half. 1064 mi t = = 2.50 hours 2nd half 426 mph We will split the calculation in a few parts: ttotal = t1st half + t2nd half = 4.95 hours

(a) What is the ground speed of the airplane for the The estimated flight time is longer than this by first and second parts of the flight ? That is, how about 18 mins. Thus we have an estimate of fast is the airplane moving relative to the ground how long the takeoff and final approach (land- (or rather, ocean) ? ing) take. 3

(c) If the takeoff took about 13 minutes, the average the field, i.e. travel at least 80 yards (73 m), resulting in acceleration during that time was touch-back ? Explain your answer (i.e a simple yes or no answer is not sufficient). Assuming Mr. Janikowski (433 − 0) knots ha i = 13 min = 0.3ms2. has picked an optimal angle for the kick (he did grad- takeoff / uate from college, after all), how long is the ball in the The average acceleration is indeed very modest. air ? What is the highest elevation of the trajectory ? Clearly, the plane’s acceleration is much higher 3. Solution on the runway (Boeing 767 typically takes off at This problem is simply asking whether the maxi- 150-200 knots), but while in the air, it reaches mum range of the ball is larger than L = 73 m. I the cruising speed relatively slowly. The average field derived the formula for the maximum range of a projec- speed during that time is tile in class, so will display the final answer here. If the 433 + 0 ball is kicked at angle θ over horizon and it lands at the hvtakeoff i = knots = 217 knots 2 same elevation, the ball travels and the aircraft covers the distance v2 sin 2θ L = . g dtakeoff = 47 nauticalmiles It immediately follows, that if v is fixed, the maximum (d) Assuming results from (c), the total time for the range 1st half of the journey is v2 Lmax = (2) (1064 − 47) mi 13 g t = + hrs = 2.57 hrs 1/2 433 mph 60 is achieved when θ = π/4 rad = 45◦. Plugging in With this in mind and wheels-up time of 6:28pm v = 30 m/s, the maximum range is PDT, I estimated that we would cross the mid- point at 9:02pm PDT. The actual time was Lmax =92m . 9:01:53 pm, and I was close enough to win the So yes, that’s a touchback. champagne. However, the uncertainties in the Now we can find the total flight time and maximum takeoff time and headwind are fairly large, so height. Both were also derived in class: even getting to within a couple of minutes of the correct time would have been good. 2v sin θ t = = 4.3 sec flight g 3. (30 points) Kickoff at the Oakland Coliseum Sebastian Janikowski is kicking off for the Oakland v2 sin2 θ Hmax = =23m Raiders. His powerful left leg launches the ball from 2g the 30-yard line with the initial velocity of 30 m/s. As- suming no air resistance, can the ball reach the end of