Physics
posted by Kelly on .
A ball is thrown upward from the top of a 50.0 m tall building. The ball's initial speed is 12.0 m/s. At the same instant, a person is running on the ground at a distance of 44.0 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?

Tr = (VfVo)/g,
Tr = (012) / 9.8 = 1.22 s. = Time to
rise to max ht.
hmax = ho + Vo*t + 0.5g*t^2,
hmax=50 + 12*1.22  4.9(1.22)^2=57.35 m
= ht. above gnd.
hmax = Vo*t + 0.5g*t^2 = 57.35 m.
0 + 4.9t^2 = 57.35,
t^2 = 11.70,
Tf = 3.42 s. = Time to fall to gnd.
d = V*Tf,
V = d/Tf = 44 / 3.42 = 12.9 m/s.=Speed
of the person.