Asked by om patel
A man throws a ball to a maximum distance of 80m.how long is it in the air and to what height does it rise? Nehlect the height of the man
Answers
Answered by
bobpursley
time in air:
hf=hi+Vi*SinTheta*timeinair-4.9 timinair^2
solving for time in air,
time in air=Vi*sinTheta/4.9
For max distance, theta=45 degrees
max distance=Vi*cosTheta*timeinair.
time in air= 80/VicosTheta
set these two time in air equal
Vi*sinTheta/4.9= 80/VicosTheta
Solve for Vi.
now solve for max height, occuring at at timeinair/2
hf=hi+Vi*SinTheta*timeinair-4.9 timinair^2
solving for time in air,
time in air=Vi*sinTheta/4.9
For max distance, theta=45 degrees
max distance=Vi*cosTheta*timeinair.
time in air= 80/VicosTheta
set these two time in air equal
Vi*sinTheta/4.9= 80/VicosTheta
Solve for Vi.
now solve for max height, occuring at at timeinair/2
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