A. Ha = Hb, ho-0.5g*t^2 = ho-(Vo*t + 0.5g*t^2).
17 - 4.9t^2 = 23.5 - (7.36*t + 4.9t^2.
17 - 4.9t^2 = 23.5 - 7.36t - 4.9t^2.
-4.9t^2 + 4.9t^2 + 7.36t = 23.5-17.
7.36t = 6.5, t = 0.883 s.
h = 17 - 4.9*(0.883)^2 = 13.2 m. Above gnd.
B. Va = Vo + g*t = 0 + 9.8*0.883 = 8.65 m/s.
Vb = Vo + g*t = 7.36 + 9.8*0.883 = 16 m/s.
Vb-Va = 16 - 8.65 = 7.35 m/s.
Somehow two students managed to open two windows at the library. Student A is 17 m above the ground and Student B is 23.5 m above the ground. At the same time that Student A drops a ball, Student B throws a ball straight down at an initial speed of 7.36 m/s.
A. At what height above the ground will Student B's ball catch up and pass Student A's ball?
B. At that moment, what is the difference in speed of the two balls?
1 answer
1 answer