Asked by Sam
A football is thrown with an initial upward velocity component of 15.0m/s and a horizontal velocity component of 18.0m/s .
How high does it get above the ground?
How much time after it is thrown does it take to return to its original height?
How far has the football traveled horizontally from its original position?
How high does it get above the ground?
How much time after it is thrown does it take to return to its original height?
How far has the football traveled horizontally from its original position?
Answers
Answered by
bobpursley
The first two questions are dependent on the vertical velocity
how high? at the top vfinal is zero.
find the time it takes to get to that point where vf is zero.
vf=vi+gt
then, using that time,
h=vi*t+1/2 gt^2 where g is -9.8
how far? use the horizontal velocity
distance=horizvel*timeinair.
how high? at the top vfinal is zero.
find the time it takes to get to that point where vf is zero.
vf=vi+gt
then, using that time,
h=vi*t+1/2 gt^2 where g is -9.8
how far? use the horizontal velocity
distance=horizvel*timeinair.
Answered by
AEC
t=1.53
complete pass time = 3.06
h= 30.4m
d(horizontal total)= 55 m/s
complete pass time = 3.06
h= 30.4m
d(horizontal total)= 55 m/s
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