A stunt driver wants to make his car jump over eight cars parked side by side below a horizontal ramp (Figure? - a stunt driver wants to make his car jump over 8 cars
A stunt driver wants the car for more than eight cars parked next to a horizontal ramp down (Fig. 3-49) beat.
Figure 3-49
(a) With what minimum speed is the horizontal distance from the ramp? The height of the ramp = 1.8 meters above the cars h and the horizontal distance from what is clear is L = 21 m.
m / s
(b) If the ramp tilted upward so that "Start angle" of 10 ° to the horizontal, the minimum speed is back?
m / s
Monday, February 8, 2010
A Stunt Driver Wants To Make His Car Jump Over 8 Cars A Stunt Driver Wants To Make His Car Jump Over Eight Cars Parked Side By Side Below A Horizontal Ramp (Figure?
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1 comments:
a) principles for resolving the kinematics of the use of (projectile motion)
Bearings to be right and below for all positive vectors.
Ignoring the effects of air resistance
Acceleration of free fall, a 9.81m/s2 =
vertical displacement, Sy = 1.8 m
Horizontal, Sx = 21m
Vertical velocity, V = (2aSy) ^ (0.5)
= (2x9.81x1.8) ^ (0.5)
= 5.94M / s
V = UY + a
0 = 5.94 + 9.81t
t = 0.606
Sx = (UX) t
21 = (UX) (0.606)
Ux = 34.7m / s (minimum speed)
b) solution into two components Ux have horizontal and vertical velocity (for example, and UY) and Part A), as you guide.thank. I hope it works
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