Asked by Dan

A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.
Answered by drwls
The time of flight is
T = 2(Vo/g)sin28

The horizontal distance is
L = (Vo^2/g)sin56

The peak height is
H = Vo^2*sin^2(28)/(2g)
Answered by sadasd
asdasd
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