Art B. answered 01/04/15
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Ph.D. Chemical Engineer Making Math, Chemistry & Physics More Fun!
Mertaug,
In order to answer these questions, we need to know a couple of physics formulas for equation of motion.
eqn (1) vf - vi = at
where vf is the final velocity
vi is the initial velocity
a is the acceleration due to gravity
t is time
eqn (2) vf2 - vi2 = 2ax
where vf, vi and a are the same as above and x = distance
On the earth, a = -32 ft/sec2 (a is negative in this case since it is in the opposite direction of the initial velocity).
On the moon, the acceleration due to gravity is 1/6 that of the earth, so a = -32/6 = -5.33 ft/sec2 (again a is negative in this case since it is in the opposite direction of the initial velocity).
Now, let's look at the questions for earth:
1) To calculate hang time, we need to determine how long it took Jake to reach his highest point (when his final velocity is 0 ft/sec). The hang time will be twice this time since it will take the same amount of time to return.
vf - vi = at
0 - 8 = -32(t)
t = 1/4 seconds
Hang time on earth= 2t = 2(1/4 seconds) = 1/2 seconds
2) To calculate maximum jump height, we use eqn (2) and solve for x:
vf2 - vi2 =2ax
0 - (8)2 = 2(-32)x
-64 = -64x
x = 1 foot on earth
Repeat for the moon:
1) Hang time:
vf - vi = at
0 - 8 = -5.33t
t = 1.5 seconds
Hang time on moon equals 2t = 2(1.5 seconds) = 3 seconds
2) Maximum Jump Height:
vf2 - vi2 = 2ax
0 - (8)2 = 2(-5.33)x
-64 = -10.66x
x = 6 feet on moon
As expected, Jake would have a longer hang time and higher jump height on the moon since the gravitational pull is less.
Hope this helps!