first ball distance = -4.9t^2 - 15.6t + 21.4
vel ball1 = -9.8t - 15.5
at ground:
0 = -4.9t^2 - 15.6t + 21.4 or 4.9t^2 + 15.6t - 21.4=0
by formula:
t = 1.035 seconds or some negative t
vel at that time = -9.8(1.035) - 15.5 = -25.64 m/s
magnitude is 25.64 m/s
2nd ball distance = -4.9t^2 + 15.6t + 21.4
repeat the steps I used for ball1
for the last part of the question, sub t = .616 into both equations and subtract their distances.
Two students are on a balcony 21.4 m above the street. One student throws a ball ver- tically downward at 15.6 m/s. At the same instant, the other student throws a ball verti- cally upward at the same speed. The second ball just misses the balcony on the way down.
What is the magnitude of the velocity of the first ball as it strikes the ground?
What is the magnitude of the velocity of the second ball as it strikes the ground?
What is the difference in the time the balls spend in the air?
How far apart are the balls 0.616 s after they are thrown?
1 answer