posted by B on .
Two students are on a balcony 18.3 m above the street. One student throws a ball (ball 1) vertically downward at 12.2 m/s; at the same instant, the other student throws a ball (ball 2) vertically upward at the same speed. The second ball just misses the balcony on the way down.
(a) What is the difference in the two ball's time in the air?
(b) What is the velocity of each ball as it strikes the ground?
(c) How far apart are the balls 0.800 s after they are thrown?
cannot seem to figure out how to find part a) i cannot get the second balls time.
a. d1 = 12.2t + 0.5gt^2 = 18.3m.
12.2t + 0.5*9.8t^2 = 18.3,
4.9t^2 + 12.2t - 18.3 = 0,
Solve using Quadratic Formula and get:
t1 = 1.054, and -3.54s.
Use positive value of t:
T1 = 1.054s = Time in air.
Vf = Vo + gt,
t2(up) = (Vf-Vo) / g,
t2(up) = (0-12.2) / -9.8 = 1.24s.
Vf^2 = Vo^2 + 2gd,
d2(up) = (Vf^2-Vo^2) / 2g,
d2(up) = (0-148.8) / -19.6 = 7.59m.
D2 = 7.59 + 18.3 = 25.89m from ground.
d = Vo*t + 0.5gt^2 = 25.89,
0 + 0.5*9.8*t^2 = 25.89,
4.9t^2 = 25.89,
t^2 = 5.28,
t(down) = 2.3s.
T2 = 1.24 + 2.3 = 3.54s in air.
T2 - T1 = 3.54 - 1.054s difference.
b. Vf^2 = Vo^2 + 2gd,
V1^2 = 12.2 + 2*9.8*18.3 = 370.88,
V1 = 19.26m/s.
V2^2 = 0 + 2*9.8*25.89 = 507.44,
V2 = 22.53m/s.
c. d1 = Vo*t + 0.5gt^2
d1 = 12.2*0.8 + 0.5*9.8*(0.8)^2 = 12.90m below the balcony.
d2=12.2*0.8 + 0.5*(-9.8)(0.8)^2 = 6.62m above the balcony
D = 6.62 + 12.90 = 19.52m apart.
A particle (a) is projected horizontally at 50m/s from the top of a tower 100m high at same instant aoother particle is projected from bottom of the tower in same vertical plain at 100m/s with elevation 60^0.after how maoy seconds will the particle collide