A man tosses a 50.0 g ball straight upward from the ground at 25m/s. at the same time a friend drops a 50.0g ball from a 20.0m tree. how long will it take each ball to hit the ground, how high will the balls rise, how much energy is possessed in each ball. at what height will the balls pass each other?
a. Balll #1: V = Vo + g*Tr = 0 @ max Ht.
Tr = -Vo/g = -25/-9.8 = 2.55 s. = Rise
time.
Tf = Tr = 2.55 s. = Fall time.
Tr+Tf = 2.55 + 2.55 = 5.10 s. = Time to
hit gnd.
Ball #2: h = 0.5g*t^2 = 20.0 m.
4.9*t^2 = 20
t^2 = 4.08
Tf = 2.02 s. = Fall time = time to hit gnd.
b. Ball #1: h = Vo*Tr + 0.5g*Tr^2 =
25*2.55 - 4.9*2.55^2 = 31.9 m. Above gnd.
Ball #2: Does not rise.
c. Ball #1: KE = PE = 0.5m*V^2 =
0.5*.05*25^2 = 15.625 J.
Ball #2: PE = KE = m*g*h = 0.05*9.8*20 =
9.8 J.
d. d1 + d2 = 20 m.
(25t-4.9t^2) + (4.9t^2) = 20
The squared terms cancel:
25t = 20
t = 0.8 s. = Time at which they pass each other.
h = 25t - 4.9t^2 = 25*0.8 - 4.9*0.8^2 =
16.9 m. = Ht. at which they pass each other.