Homework Help: physic

Posted by Anonymous on Wednesday, August 29, 2012 at 9:54pm.

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?

physic - Henry, Friday, August 31, 2012 at 10:19pm
Tr = (Vf-Vo)/g = (0-12) / -9.8 = 1.22 s=
Rise time or time to reach max .ht.

hmax = ho + (Vf^2-Vo^2)/2g.
hmax = 24.4 + (0-144) / -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.

Answer this Question

Related Questions

Physics - A ball is thrown upward from the top of a 24.6 m tall building. The ...
physics - A ball is thrown upward from the top of a 24.4-m-tall building. The ...
Physics - A ball is thrown upward from the top of a 50.0 m tall building. The ...
physics - A ball is thrown upward from the top of a 40.0 m tall ...
physics - A ball is thrown upward with an initial velocity of 6.0m/s by someone ...
physics - A ball is thrown from the top of a building upward at an angle of 66 &...
Math: Word Problem - A ball is thrown upward from the top of a building which is...
Physics - The ball is thrown upward with a speed of 38.0m/s from the top of a ...
phusics - A ball is thrown vertically upward from top of a building with speed ...
physics - Assume acceleration due to gravity=10 m/s. A ball is thrown ...

For Further Reading

First Name:
School Subject:
Answer: