What is the magnitude of g at a height above Earth's surface where free-fall acceleration equals 6.5m/s^2?
1 answer:
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Acceleration is given by:
where
v is the final velocity
u is the initial velocity
t is the time interval
Let's apply the formula to the different parts of the problem:
A)
Let's convert the quantities into SI units first:
t = 4.0 min = 240 s
So the acceleration is
B)
As before, let's convert the quantities into SI units first:
t = 94 s
So the acceleration is
C)
For this part we have to use a different formula:
where we have
v = 0 is the final velocity
u = 89.2 m/s is the initial velocity
a is the acceleration
d = 75 m is the distance covered
Solving for a, we find