physic
posted by Anonymous .
A ball is thrown upward from the top of a 24.4 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 31.4 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 = (012) / 9.8 = 1.22 s=
Rise time or time to reach max .ht.
hmax = ho + (Vf^2Vo^2)/2g.
hmax = 24.4 + (0144) / 19.6 = 31.75 m=
ht. above gnd.
h = Vo*t + 0.5g*t^2 = 31.75 m.
0 + 4.9t^2 = 31.75
t^2 = 6.48
Tf = 2.55 s. = Time to fall to gnd.
D = r*(Tr+Tf).
r = D/(Tr+Tf)
r=31.4 / (1.22+2.55) = 8.33 m/s.
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