Asked by nour
suppose a ball thrown downward with initial velocity 3 m/s instead of being dropped
a)what then would be its position after 1s and 2s ?
b)what would its speed be after 1 s and 2 s ?
a)what then would be its position after 1s and 2s ?
b)what would its speed be after 1 s and 2 s ?
Answers
Answered by
Henry
a. d = Vo*t + 0.5g*t^2,
d=3*1 + 4.9*1^2 = 7.9m below starting point.
d = 3*2 + 4.9*2^2 = 25.6m
b. Vf^2 = V0^2 + 2g*d,
Vf^2 = 3^2 + 19.6*7.9 = 163.84,
Vf = 12.8m/s.
Vf^2 = 3^2 + 19.6*25.6 = 510.76,
Vf = 22.6m.
d=3*1 + 4.9*1^2 = 7.9m below starting point.
d = 3*2 + 4.9*2^2 = 25.6m
b. Vf^2 = V0^2 + 2g*d,
Vf^2 = 3^2 + 19.6*7.9 = 163.84,
Vf = 12.8m/s.
Vf^2 = 3^2 + 19.6*25.6 = 510.76,
Vf = 22.6m.
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.