Either 175 N or 157 N depending upon how the value of 48° was measured from.
You didn't mention if the angle of 48° is from the lug wrench itself, or if it's from the normal to the lug wrench. So I'll solve for both cases and you'll need to select the desired answer.
Since we need a torque of 55 N·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 N·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.
Ideally, the force will have a 0° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".
If the angle of 48° is from the normal to the lug wrench, the usefully applied power will be:
U = F*cos(48)
where
U = Useful force
F = Force applied
So solving for F and calculating gives:
U = F*cos(48)
U/cos(48) = F
117 N/0.669130606 = F
174.8537563 N = F
So 175 Newtons of force is required in this situation.
If the 48° is from the lug wrench itself, that means that the force is 90° - 48° = 42° from the normal. So doing the calculation again (this time from where we started plugging in values) we get
U/cos(42) = F
117/0.743144825 = F
157.4390294 = F
Or 157 Newtons is required for this case.
The reasoning for this is false
<span>Her center of mass will rise 3.7 meters.
First, let's calculate how long it takes to reach the peak. Just divide by the local gravitational acceleration, so
8.5 m / 9.8 m/s^2 = 0.867346939 s
And the distance a object under constant acceleration travels is
d = 0.5 A T^2
Substituting known values, gives
d = 0.5 9.8 m/s^2 (0.867346939 s)^2
d = 4.9 m/s^2 * 0.752290712 s^2
d = 3.68622449 m
Rounded to 2 significant figures gives 3.7 meters.
Note, that 3.7 meters is how much higher her center of mass will rise after leaving the trampoline. It does not specify how far above the trampoline the lowest part of her body will reach. For instance, she could be in an upright position upon leaving the trampoline with her feet about 1 meter below her center of mass. And during the accent, she could tuck, roll, or otherwise change her orientation so she's horizontal at her peak altitude and the lowest part of her body being a decimeter or so below her center of mass. So it would look like she jumped almost a meter higher than 3.7 meters.</span>
