A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.? - jump de teck deck
What is the minimum speed must be disposed horixzontal ramp? The ramp height of 1.5 m above the cars, and the horizontal distance to be clear, is 20m. If the slide tilted upward so that the start angle of 10 degrees above the horizontal, the minimum speed is back?
Friday, November 6, 2009
Jump De Teck Deck A Stunt Driver Wants To Make His Car Jump Over 8 Cars Parked Side By Side Below A Horizontal Ramp.?
2 comments:
Let's see how long it takes to pass up to 1.5 meters.
Of the vertical angle:
Viv first vertical velocity = 0 m / s (from the ramp is horizontal)
the initial level of Tu = 1.5 m
final height D = 0
Acceleration = -9.8 m / s ^ 2
The application of the equation of the distance
1/2at = d ^ 2 + VT + Di
0 = -4.9 t ^ 2 + 1.5
t = sqrt (1.5/4.9) = 0.55
Among the horizontal angle: The car must be 20 meters in 0.55 speed for the least 20m/0.55s Cross = 36,4 m / s
If the ramp is inclined at 10 degrees. Speed components now vertically and horizontally:
Viv = sin (10) Vi
HIV = cos (10) Vi
Of the vertical angle:
Viv first vertical velocity = sin (10) Vi m / s
the initial level of Tu = 1.5 m
final height D = 0
Acceleration = -9.8 m / s ^ 2
The application of the equation of the distance
1/2at = d ^ 2 + VT + Di
0 = -4.9 t ^ 2 + sin (10) VI.TAR 1.5
Among the horizontal angle: The car has 0.55 speed in the 20m cross so that the minimum
cos (10) xt Vi = 20 m
VI = 20 / (t * cos (10))
Replacing the first equation yields
0 = -4.9 t ^ 2 + sin (10) .20 / (T * cos (10)). T + 1.5
0 = -4.9 t ^ 2 + 20sin (10) / (COS (10)) + 1.5
0 = -4.9 t ^ 2 + 3.52 + 1.5
0 = -4.9 t ^ 2 + 5.02
t = sqrt (5.02/4.9) = 1.01 s
and since Vi = 20 / (t * cos (10))
(1.01 Vi = 20 / cos (10)) = 20.06 m / s, which is substantially less than 36 m / s.
Let's see how long it takes to pass up to 1.5 meters.
Of the vertical angle:
Viv first vertical velocity = 0 m / s (from the ramp is horizontal)
the initial level of Tu = 1.5 m
final height D = 0
Acceleration = -9.8 m / s ^ 2
The application of the equation of the distance
1/2at = d ^ 2 + VT + Di
0 = -4.9 t ^ 2 + 1.5
t = sqrt (1.5/4.9) = 0.55
Among the horizontal angle: The car must be 20 meters in 0.55 speed for the least 20m/0.55s Cross = 36,4 m / s
If the ramp is inclined at 10 degrees. Speed components now vertically and horizontally:
Viv = sin (10) Vi
HIV = cos (10) Vi
Of the vertical angle:
Viv first vertical velocity = sin (10) Vi m / s
the initial level of Tu = 1.5 m
final height D = 0
Acceleration = -9.8 m / s ^ 2
The application of the equation of the distance
1/2at = d ^ 2 + VT + Di
0 = -4.9 t ^ 2 + sin (10) VI.TAR 1.5
Among the horizontal angle: The car has 0.55 speed in the 20m cross so that the minimum
cos (10) xt Vi = 20 m
VI = 20 / (t * cos (10))
Replacing the first equation yields
0 = -4.9 t ^ 2 + sin (10) .20 / (T * cos (10)). T + 1.5
0 = -4.9 t ^ 2 + 20sin (10) / (COS (10)) + 1.5
0 = -4.9 t ^ 2 + 3.52 + 1.5
0 = -4.9 t ^ 2 + 5.02
t = sqrt (5.02/4.9) = 1.01 s
and since Vi = 20 / (t * cos (10))
(1.01 Vi = 20 / cos (10)) = 20.06 m / s, which is substantially less than 36 m / s.
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