Answer:
Both forces are the same.
Explanation:
The problem states that:
The force exerted on the moon by earth is given by:
where K is the gravitational constant, and d is the separation distance between the earth and the moon.
The force exerted on the earth by moon is given by:
where K is the gravitational constant, and d is the separation distance between the earth and the moon.
The relation is therefore:
As you can see, they are the same.
(a) 30.9 m
Let's analyze the motion of the parachutist. Its vertical position above the ground is given by
where
h = 45 m is the initial height
u = 10 m/s is the initial velocity (upward)
t is the time
g = -9.8 m/s^2 is the acceleration of gravity (downward)
Substituting t=3 s , we find the height of the parachutist when it opens the parachute:
(b) 44.1 m
Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by
y = h + ut
where
h = 45 m is the initial height
u = 10 m/s is the initial velocity (upward)
t is the time
Substituting t = 3 s, we find
y = 45 m + (10 m/s)(3 s) = 75 m
So the distance between the balloon and the parachutist after 3 s is
d = 75 m - 30.9 m = 44.1 m
(c) 8.2 m/s downward
The velocity of the parachutist at the moment he opens the parachute is:
v = u +gt
where
u = 10 m/s is the initial velocity (upward)
t is the time
g = -9.8 m/s^2 is the acceleration of gravity (downward)
Substituting t = 3 s,
v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s
where the negative sign means it is downward
After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of
a =+5 m/s^2
where we put a positive sign since this time the acceleration is upward.
The total distance he still has to cover till the ground is
d = 30.9 m
So we can find the final velocity by using
where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:
a = -5 m/s^2
Solving for v,
(d) 5.24 s
We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using
where
a = -5 m/s^2 is the deceleration
v = 8.2 m/s is the final velocity
u = 19.4 m/s is the initial velocity
t is the time
Solving for t, we find
And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is
t = 3 s + 2.24 s = 5.24 s