Asked by Anthony
A Football is kicked 66.6 meters in 5.39 seconds. What was the initial velocity it was kicked?
magnitude and direction above horizontal
magnitude and direction above horizontal
Answers
Answered by
Jennifer
x = vx*t
y = vy*t - 0.5*9.8*t^2
Assuming it hits the ground at 5.39 seconds:
66 = vx*5.39
0 = vy * 5.39 - 0.5*9.8*5.39^2
Where vx is the speed in the x direction, vy is the speed in the y direction.
The magnitude of the speed is
(vx^2 + vy^2)^0.5
The direction of the speed is given by
tan(x) = vy / vx
Solve for the angle x
y = vy*t - 0.5*9.8*t^2
Assuming it hits the ground at 5.39 seconds:
66 = vx*5.39
0 = vy * 5.39 - 0.5*9.8*5.39^2
Where vx is the speed in the x direction, vy is the speed in the y direction.
The magnitude of the speed is
(vx^2 + vy^2)^0.5
The direction of the speed is given by
tan(x) = vy / vx
Solve for the angle x
Answered by
Anthony
how do you get the vy by it self when trying to get tionhe speed in the y direct
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