Asked by Angel
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?
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?
Answers
Answered by
Henry
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.
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.
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