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

Answer 1

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