Posted by **Jenn** on Sunday, September 23, 2012 at 3:07pm.

A ball is thrown upward from the top of a 24.6 m tall building. The ball's initial speed is 12 m/s. At the same instant, a person is running on the ground at a distance of 31.2 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, September 24, 2012 at 4:22pm
hmax = ho + (V^2-Vo^2)/2g.

hmax = 24.6 + (0-144)/-19.6 = 31.95 m

Above gnd.

hmax = Vo*t + 0.5g*t^2 = 31.95 m.

0 + 4.9t^2 = 31.95

t^2 = 6.52

Tf = 2.55 s. = Fall time.

d = V*Tf = 31.2 m.

V * 2.55 = 31.2

V = 12.22 m/s.

- Physics -
**Henry**, Monday, September 24, 2012 at 4:43pm
Correction:

Tr = (V-Vo)/g = (0-12)/-9.8 = 1.22 s. =

Rise time.

hmax and Tf are correct as shown in the

previous analysis.

d = V * (Tr+Tf) = 31.2 m.

V * (1.22+2.55) = 31.2

V = 8.28 m/s.

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