0.65 s after leaving dock, vertical velocity component is
Vy = gt = 6.37 m/s
Vertical displacement will be
y = (1/2)gt^2 = 2.07 m
Horizontal displacement must be
X = sqrt[8.4^2 - (2.07)^2] = 8.14 m
Horizontal velocity component must be
Vx = X/0.65s = 12.52 m/s
That is also the speed as it leaves the edge of dock
A child operating a radio-controlled model car on a dock accidentally steers it off the edge. The car's displacement 0.65 s after leaving the dock has a magnitude of 8.4 m. What is the car's speed at the instant it drives off the edge of the dock?
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