Asked by Anonymous
A flowerpot falls from a windowsill 33.0 m above the sidewalk.
a) How much times does a passerby on the sidewalk below have to move out of the way before the flowerpot
hits the ground?
b) What is the flowerpot's velocity when it strikes the ground?
a) How much times does a passerby on the sidewalk below have to move out of the way before the flowerpot
hits the ground?
b) What is the flowerpot's velocity when it strikes the ground?
Answers
Answered by
Damon
g = a = -9.81 m/s^2
v = 0 - 9.81 t
h = 33.0 + 0 t - 4.9 t^2
h = 0 at crash
4.9 t^2 = 33
t = 2.6 seconds to disaster
back to v = 0 -9.81 t
= - 9.81*2.6 , negative because downward like g
v = 0 - 9.81 t
h = 33.0 + 0 t - 4.9 t^2
h = 0 at crash
4.9 t^2 = 33
t = 2.6 seconds to disaster
back to v = 0 -9.81 t
= - 9.81*2.6 , negative because downward like g
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.