Asked by nelo
a daredevil on a motorcycle leaves the ramp with a speed of 35.0 miles per hour. if his speed is 33.0 miles per hour when he reaches the peak of his path, what is the maximum height that he reaches. ignore friction and air resistance
Answers
Answered by
drwls
His "launch angle" is
theta = cos^-1(33/35) = 19.46 degrees
The initial vertical velocity component is
Voy = 35*sin(19.46)
= 11.66 mph
= 17.1 ft/s
The time to reach maximum height is
t' = (Voy)/g = 0.53 seconds
Maximum height = (1/2)*Voy*t' = 4.5 feet PLUS the initial ramp height where he leaves it. They should have told you what that is.
theta = cos^-1(33/35) = 19.46 degrees
The initial vertical velocity component is
Voy = 35*sin(19.46)
= 11.66 mph
= 17.1 ft/s
The time to reach maximum height is
t' = (Voy)/g = 0.53 seconds
Maximum height = (1/2)*Voy*t' = 4.5 feet PLUS the initial ramp height where he leaves it. They should have told you what that is.
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