A car drives straight off the edge of a cliff that is 59 m high.?
84432
Assuming this occurs in a vacuum:\nvt = 140m, where t = time, v = speed of car\n1/2 * at^2 = 59m\n\nSolving:\nt = sqrt( 2 * 59/9.8 )\nv = 140/t = 40.3459615100m/second\n\nFor extra points (from your prof?):\nTake into account the change in acceleration due to gravity.  Assume the car leaves the top of the cliff (a nice sheer edge a la Wiley Coyote) at 9.8 m/s^2 and descends 59 m.  Using F = G*mc*me/r^2, where F = acceleration force, G = gravitational constant (assume 6.67E-11), mc = mass of car, me = mass of earth (assume 5.972E24), r = radius.  Based on F = ma (classical formula) and a mass of car = 1000 kgs, then the radius, r, will change from 6377441.301 to 6,377,382.30 meteres (a descent of 59 m) which will increase acceleration due to gravity from 9.8 to 9.80018 m.  If we recalculate velocity of the car using this final acceleration due to gravity instead of 9.8 then car speed will have been 40.34633 m/s instead of 40.34596 m/s (rounded to 5 decimal points), a difference of -0.00093%.  Therefore the speed of the car would be somewhere between 40.34633 m/s and 40.34596 m/s.  Higher accuracy would be obtained had we known the mass of the car and elevation of the cliff plus we would alter the equation to take into account the change in acceleration due to gravity as the car plunged 59 m.  Additional accuracy would be to take into account acceleration due to centrifugal force; we would need to know the cardinal direction of the car's plunge.
494819
MORE OR LESS 135 KM/H.
784055
The way to do this is to first work out the acceleration required to turn the car round the curve.\nFrom there it is a simple matter of working out what angle the road needs to be at so that gravity will provide that acceleration.
968172
Beat the helicopters? In a car? LOL......... I don't think so unless it's on the Bonneville Salt Flats in Utah... AND you are driving something that sets land speed records..........LOL....... beat the helicopters........LMAO
82342
This is an impossibility.  Cars don't travel at the speed of light.
149924
it will hit the back of  the car
684454
Let's consider the case when the Car is right in front.\n\nAt that instance, the car is travelling tangentially to a circle of radius 132ft, where the Camera is the centre of the circle.\n\nCircumference of the Circle = 2*(pi)*R\n= 2 * pi * 132 = 264*(pi) ft.\n\nTime taken to travel the entire circumference by the car\nT = circumference/velocity = 264*(pi)/264 = pi seconds\n\nHence the car would require pi s to cover 360 degrees of the circle.360 degrees = 2(pi) radians.\n\nW = angular velocity = 2(pi)/pi = 2 radians/sec.\n\ncan't help about half a sec later now
743173
Its a Maserati quattroporte
861510
The way to solve this problem is to decompose the forces acting on the car, but not in the x and y direction. Instead, decompose the forces in the direction of the incline and the normal force.\n\nHence, we have:\n\nF(friction) = μN, moving up the incline, and μ is the coefficient of static friction\n\nThen we have gravity acting on the car, and the component that's acting down the incline is = mgsinθ, where θ is the incline of 15°.\n\nFor the car to be safe in making this turn, the forces along the incline must be equal:\n\n(a) F(friction) = μN = mgsinθ = (1450)(9.8)sin(15°) = 3677.82 N\n\nBut the forces acting perpendicular to the incline on the car must also be zero:\n\nN = mgcosθ\n\nNow substituting this into the previous equation, we get:\n\nμN = μ(mgcosθ) = mgsinθ, this implies that the coefficient of static friction is:\n\n(b) μ = sinθ/cosθ = tanθ = tan(15°) = 0.268
59101
It actually would be possible, yes.  The momntum of the car doesn't really change when it drives onto the ramp.  if the ramp is smooth enough then the tires will just spin and the car won't launch through the roof.  If the ramp has enough traction then this trick would be hard on the transmission would probabaly kill a stick shift, but again the car would not launch through the roof.
86689
The the car would stall
845823
the speeds are too small, so classical mechanics should be fine:\n\nthe speed of the car is 45-10 = 35mph
413567
By getting everyone out of the car and then pushing it off a very high cliff.
1163276
The angle the vehicle was traveling
997426
On top of a police car.
908785
very close to 2 meters.
115023
Imagine a `speed` x `time` chart. Plotting the data, we will have parallelogram whose area will the length the car walked between the two speeds.\n\nWe'll have, then, 30*((5.56+13.89)/2) = 291.75m