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Today I was driving on a hill side and on the opposite lane this very careless gentleman was traveling at a very high speed on a large truck that I surely thought his truck would of lifted off the ground. I started to think about how one would even calculate at what speed a certain mass would need to be traveling in order to launch itself from a cliff that makes an angle $\theta$ and how far that object would land.

It seemed like a fairly easy problem at first, so I drew a free body diagram with all (to my knowledge) possible variables present and how chancing them would affect the results. However I am unsure where to start.

Diagram

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  • $\begingroup$ Your diagram should use the convention of direction of travel towards the right as that is what most brains will see - regardless of arrows. This is probably what mislead @Симон Тыран. $\endgroup$ – Russell McMahon Jul 11 '15 at 3:29
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The car becomes "weightless" when the curvature of the road is sufficient that the car does not stay connected to the road. Angles don't really matter - what matters is a change in the direction of the road.

As you know, an object going around in a circle needs a force $F = \frac{mv^2}{r}$ to stay in the orbit at radius $r$. Normally, when you drive over a bump the force of gravity is sufficient to keep you connected - but now we can see that the car will lift off if the radius of curvature $r$ is too small:

$$\frac{mv^2}{r} \gt mg\\ r \lt \frac{v^2}{g}$$

For a car going 100 km/h, that puts the limiting radius at about 80 m. Another way of thinking about this: if you think of the car driving off a cliff, it would drop 5 m in the first second. If the road drops away faster than that, the car will appear to lift off. That's perhaps more intuitive than the radius of curvature.

enter image description here

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    $\begingroup$ As an aside; the primary force keeping the tires on the ground is actually the suspension system on the vehicle; in many cases a vehicle should experience a separation from the ground, such as going over a pothole or a small bump at high speed, but the suspension springs push the wheels down faster than gravity pulls the car down, which helps "fill in the gap" so that the vehicle maintains traction. The calculation would contain extra terms for suspended vehicles vs. suspensionless vehicles (such as bicycles or train cars). $\endgroup$ – Asher Jul 11 '15 at 4:27
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First I will make a few assumptions: the hill is flat at the its top (so its surface is perpendicular to gravity); the side of the hill is flat and makes an angle $\theta$ relative to the top; the end of the hill is sufficiently far away such that it will never be reached during the "jump" sideways of the top of the hill; the tradition from the top to the side of the hill is sudden/a sharp corner; the car is a point mass which initially slides without friction over the top of the hill at a constant velocity $v$; atmospheric forces can be neglected.

These assumptions are a little bit like the spherical cow, for example the transition from the top to the side of the hill will most likely not be a sharp edge, but more gradually and the car will have a suspension that will keep its wheels in contact with the ground for longer. Using these assumptions then the car will always "lift off" at the edge of the top of the hill, but the distance airborne will depend on $v$. When looking at the path of the car I will define the horizontal motion as the x-axis and the vertical motion as the y-axis. The position and time at the moment the car reaches the edge between the top and the side of the hill will all be defined as zero. Along the x-axis there are no forces acting on the car and the initial velocity in that direction is $v$. Along the y-axis gravity is acting downwards at an acceleration $g$ and the initial velocity is zero. As long as the car does not make contact again with the side of the hill, then the motion in the x and y directions can be described with,

$$ x(t) = v t, $$

$$ y(t) = -\frac{1}{2} g t^2. $$

This motion along the x direction can be used to find the rate at which the surface drops below the car, namely at a given $x$ the corresponding y coordinate of the surface is,

$$ y_s = -x \tan\theta = -v t \tan\theta. $$

The car will make contact with the side of the hill again at $t_c$, when the difference between $y(t_c)$ and $y_s$ is zero again ($t_c>0$),

$$ y(t_c) - y_s = -\frac{1}{2} g t_c^2 + v t_c \tan\theta = 0, $$

$$ t_c = \frac{2 v \tan\theta}{g}. $$

The distance airborne along the side of the hill can be calculated with,

$$ d = \sqrt{x(t_c)^2 + y(t_c)^2} = \frac{2 v^2 \tan\theta}{g} \sqrt{1 + \tan^2\theta}. $$

If this distance is much larger than the distance it takes to transition from the top to the side of the hill, then it will be very likely that the car will lose contact with the ground. If that is not the case than it is harder to find out without knowing more about how the top transitions in to the side of the hill and what kind of suspension the car has.

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Define the origin of your coordinate system at the top of the hill with the vehicle travelling in the $+x$ direction with initial velocity $v$ and initial position $s=(0,0)$. The road then "falls away" from the vehicle at a rate of $v \tan(\theta)$, so the altitude $h$ of the vehicle above the downward sloping road can then be written as a function of time $t$:

$h=v \tan(\theta)t - {\frac{gt^2}{2}}$.

That is simply the difference between the descent of the road (first term) and the descent of the vehicle due to gravity (second term). To find the time of impact, set $h=0$ and solve for $t$, which gives

$t{_{impact}}=\frac{2v \tan(\theta)}{g}$ .

This distance $s{_x}$ traveled in the $x$ direction prior to impact is then

$s{_x}=vt{_{impact}}$.

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It just oblique projectile motion U= speed of truck h= height of hill If you want to calculate how far from hill the truck land after flight use U multiply by the whole root of 2H/g g is gravitational acceleration =10m/s2

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protected by Qmechanic Jul 11 '15 at 7:27

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