Asked by Uzz
A juggler throws a ball straight up into the air with a speed of 11 .
With what speed would she need to throw a second ball half a second later, starting from the same position as the first, in order to hit the first ball at the top of its trajectory?
With what speed would she need to throw a second ball half a second later, starting from the same position as the first, in order to hit the first ball at the top of its trajectory?
Answers
Answered by
Henry
h = (Vf^2 - Vo^2 / 2g,
h = (0 - (11)^2) / -19.6 = 6.17m,
t(up) = (Vf - Vo) / g,
t(up) = (0 - 11) / -9.8 = 1.12s.
t = 1.12 - 0.5 = 0.62s.
Vo*t + 0.5gt^2 = 6.17m,
0.62Vo - 4.9*(0.62)^2 = 6.17,
0.62Vo - 1.88 = 6.17,
0.62Vo = 6.17 + 1.88 = 8.05,
Vo =12.98m/s.
h = (0 - (11)^2) / -19.6 = 6.17m,
t(up) = (Vf - Vo) / g,
t(up) = (0 - 11) / -9.8 = 1.12s.
t = 1.12 - 0.5 = 0.62s.
Vo*t + 0.5gt^2 = 6.17m,
0.62Vo - 4.9*(0.62)^2 = 6.17,
0.62Vo - 1.88 = 6.17,
0.62Vo = 6.17 + 1.88 = 8.05,
Vo =12.98m/s.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.