On January 16, 2003, 81.7 seconds into the ascent, a piece of foam, weighing abo
ID: 2971726 • Letter: O
Question
On January 16, 2003, 81.7 seconds into the ascent, a piece of foam, weighing about 1.7 pounds, and being roughly the size, shape and weight of a large loaf of bread, separated from the bipod ramp of the space shuttle Columbia's external fuel tank and impacted the leading edge of the shuttle's left wing. At that time the shuttle was traveling at approximately 1,568 mph (Mach 2.46) and was at an altitude of 66,000 feet. Based on film evidence, the foam traveled the 58-foot distance from the ramp to the wing in 0.16 seconds. Assuming a constant acceleration relative to the space shuttle of [ ? ]feet per second squared, the foam hit the wing at a speed of [ ? ] feet per second, or [ ? ] miles per hour. Unbeknownst to the astronauts and observers on earth, it struck the wing hard enough to cause the destruction of the Columbia two weeks later during re-entry on February 1, 2003. Note: A much more involved calculation estimated the speed of the impact to be about 530 mph. You can find very detailed information about all aspects of the Columbia accident in this official report. Your answer to be within one tenth of one percent of what it considers the true answer. Calculate your answers based on the given data and assumptions to at least four digits.Explanation / Answer
Relative to the shuttle simply means that the piece of foam and the shuttle may be considered as being at rest since they are traveling upward together until the foam breaks off.
Under constant acceleration:
s = (1/2)a*t^2
56 = (1/2)a*0.16^2
a = 2*56/0.16^2 = 4531.25 ft/sec^2
v = a*t = 4531.25*0.16 = 725 ft/sec
What you calculated is the average speed of the foam.
Under constant acceleration: Vavg = (Vstart + Vfinal)/2 = Vfinal/2 = 362.5 ft/sec
So Vfinal = 2*362.5 = 725 ft/sec
This speed is relative to the shuttle and directed toward the rear of the shuttle. Relative to the Earth, the foam is still traveling upward.
A little more info I suppose.
v = v0 + a*t ,,,,,, constant acceleration with an initial velocity of v0
But since v0 = 0 we have v = a*t
And by definition v = ds/dt so ds/dt = a*t and ds = a*t*dt
Integrate to get s = (1/2)a*t^2