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 = (Vf-Vo)/g,

Tr = (0-12) / -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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