My turn to give questions. A motorbike is at rest 5m away from a 10m long ramp at a 30 degree angle. An identical ramp is set up facing the opposite direction with a 20m gap between the two ramps. If the motorbike provides 4900N of force while in contact with the ground and has a mass of 700kg, will it make the jump? Ignore friction, air resistance, whatever else. Assume this is on Earth and not a planet with different gravity.
Good ol' mechanics
a = 7m/s
2v
2 =2*a*r
v = (2*a*r)
1/2r = 15
v = ~14.5 (nearest 1 dp)
v
y = ~7.25
Assuming that the height of the adjacent ramp is the SAME as the height of the previous ramp (ie: We don't even need to take the height of the ramp into account)
Because we don't need to take the height into account, then the time it takes to get to maximum height is the same as it takes to get back down to minimum (ie when v = 0) therefore
0 = u - (1/2)at
t = 2u/a
t = 1.48 seconds (nearest 2dp)
In this time his horizontal velocity is given as 12.56 m/s (nearest 2dp)
therefore he traveled 18.56 m, and therefore didn't make the ramp.
If the ramp wasn't there and we assume the sides of the canyons are on the same level then we can take our origin to be the ground and thus the person actually started from h meters high
h = sin(30)*10
h = 5
0 = 5 + ut - 1/2*a*t^2
Using quadratic formula t = 1.911
meaning he would travel 25 meters, which would clear the gap!
Ok, say you had an aerial vehicle, say a commercial jet
and you had a double-barred cannon built into the aircraft. (let's go with the Hull)
If the Jet was over 2.5 tons (5,000 Lbs) and was falling at 1400 m/s at a 65º angle facing earth
plus the applied force of gravity 9.8 m/s^2.
Including we know the cannon's both barrels fire at the same time.
The cannon shoots at a force of 240 N.
Then how many shots are needed to slow the jet to a safe landing?
The jet low enough to crashland and be destroyed, about 130,000 M up.
2500 kg - 240 N - 9.8 m/s^2 - 1400 m/s
I hate to break it to you, but the force due to gravity would be 2500*9.8 N which is about 25000N. A 240N cannon won't do very much to slow it down.
Note: when you say falling, I'm assuming that you mean that no force is being provided by the engines. Also, I'm of course ignoring air resistance here, so that might change things.
He's generally right, if your cannon can only apply 240 N, this would be insufficient (because we're working with acceleration, it doesn't matter how many times your cannon is firing, a cannon firing at 240 N would apply a constant deceleration, which opposed to the constant acceleration of gravity, would not stop the plane from falling). However, lets for the purpose of argument see how much force it would take to decelerate a cannon completely so that it landed safely (by landing safely we assume that means it is not accelerating when it touches the ground).
So first things first, we need to know the actual acceleration that would be required to successfully land on the ground, we're assuming the cannon is placed on the planes center of mass and air resistance is neglected (If we didn't assume the cannon was in the plane's center of mass, then we'd need a length of the plane. As well, if we assumed that air resistance wasn't neglected, than we need the temperature of the surrounding area (and if it wasn't constant we'd need a function to describe how it changes with altitude)).
The plane is falling so it's y velocity is given as -1400 m/s or we'll call it -u, the planes mass is 2.5 tons (we'll call m) and the force due to the cannon firing would be F and gravity would be g, we'll also call d the distance the plane is from the sky and the ground (this is 130km up, so the plane is actually a space shuttle). For the moment, we're also going to assume g is constant, but this assumption isn't technically correct because the space shuttle is very high.
the equation for velocity is in the y direction would be
v = -u - at
a is given as (g - F/m) therefore the velocity would be given as
0 = -u - (g - F/m)t
this means that t = -u / (g - F/m)
Therefore, in order for t to be positive, F / m needs to be greater than g, which makes sense.
Now that we've got one version of t, we need to equate it for another version of t, lets consider this equation
y = y1 + u1t + 1/2(a)(t)^2
subbing in our values we have
0 = d - ut - 1/2 (g - F/m) * t^2
We then sub in for t
0 = d + u
2 / (g - F/m) - 1/2*u
2 / (g - F / m)
Huh would you look at that, the denominators of the first two terms are the same, that's nice
d = -u
2 / 2(g - F/m)
2*d(g - F/m) = -u
22*d*F/m = (u
2 + 2*dg)
F = (u
2 + 2*dg)*m / 2*d
Therefore subbing all our values in we get
F = 43,346 (to the nearest dp)
Now F is the required force for the plane to maintain it's altitude, if we assume again that the plane doesn't change angle, then the amount of force the cannon actually has to release is
F
t * sin(65) = F
F
t = F / sin(65)
this is 47,827 N (to the nearest integer)
This was done pretty quickly so there may of been a few mistakes. Also this assumed that air resistance was negligable, no torque was applied to the plane and that gravity was constant.
EDIT: Realizing now you can probably do it an easier way by solving the equation v
2 = u
2 + 2ar
EDIT2: Found out I missed a -, fixed now
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