Posted by Kelly on Sunday, February 5, 2012 at 12:36pm.
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?

Physics  Henry, Monday, February 6, 2012 at 7:53pm
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.
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