Asked by Marvin
a stone is thrown upward from the ground with a velocity of 80 m/s. how long will it take to reach a point 3 m above the ground on its way down?
Answers
Answered by
Elena
The max height is
H= v²/2g =80²/2•9.8 = 326.53 m.
The time of upward motion is
t1 =v/g =80/9.8=8.16 s,
the time of downward motion is
t2=sqrt{2(H-h)/g}=sqrt{2(326.53-3)/9.8} =8.126.
The total time is
t=t1+t2 =8.16+8.126 =16.126 s.
H= v²/2g =80²/2•9.8 = 326.53 m.
The time of upward motion is
t1 =v/g =80/9.8=8.16 s,
the time of downward motion is
t2=sqrt{2(H-h)/g}=sqrt{2(326.53-3)/9.8} =8.126.
The total time is
t=t1+t2 =8.16+8.126 =16.126 s.
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