# physics

posted by .

Two students are on a balcony 24.4 m above the street. One student throws a ball, b1, vertically downward at 20.3 m/s. At the same instant, the other student throws a ball, b2, vertically upward at the same speed. The second ball just misses the balcony on the way down.
(a) What is the difference in time the balls spend in the air?

(b) What is the velocity of each ball as it strikes the ground?
velocity for b1

velocity for b2

(c) How far apart are the balls 0.510 s after they are thrown?

• physics -

The First Ball:
d = Vo*t + 0.5g*t^2 = 24.4 m.
20.3t + 4.9t^2 = 24.4
4.9t^2 + 20.3t - 24.4 = 0
Tf = 0.973 s.=Fall time or time in air.

The 2nd Ball:
Tr = (V-Vo)/g = (0-20.3)/-9.8 = 2.07 s.=
Rise time.
h = ho + Vo*t + 0.5g*t^2
h=24.4 + 20.3*2.07 - 4.9(2.07)^2=45.4 m
Above gnd.
d = Vo*t + 0.5g*t^2 = 45.4 m.
0 + 4.9t^2 = 45.4
t^2 = 9.27
Tf = 3.04 s.
Tr + Tf = 2.07 + 3.04 = 5.1 s. = Time
in air.

a. 5.1 - 0.973 = 4.14 s.

b. V1^2 = Vo^2 + 2g*d.
V1^2 = (20.3)^2 + 19.6*24.4 = 890.3
V1 = 29.8 m/s.

V2^2 = 0 + 19.6*45.4 = 889.8
V2 = 29.8 m/s.

c. h1 = ho - (Vo*t + o.5g*t^2.)
h1 = 24.4 - (20.3*0.51 + 4.9(0.51)^2
h1 = 24.4 - 11.63 = 12.8 m. Above gnd.

h2 = ho + (20.3*0.51 - 4.9(0.510^2
h2 = 24.4 + 9.08 = 33.5 m. Above gnd.

Da = h2 - h1 = 33.5 - 12.8 = 20.7 m.
Apart.