Asked by Anonymous
A baseball is thrown horizontally from a height of 11.20 m above the ground with a speed of 27.5 m/s. Where is the ball after 0.90 s have elapsed?
The ball is 1 a horizontal distance of 2 m from the launch point.
The ball is 1 a horizontal distance of 2 m from the launch point.
Answers
Answered by
Henry
Xo = 27.5 m/s.
Yo = 0.
T = 0.9 s. = Time in flight.
Dx = Xo*T = 27.5m/s * 0.9s = 24.75 m.
from launch point.
h = ho _ 0.5g*T^2,
h=11.2 - 4.9*(0.9)^2 =11.2-3.97=7.2 m.
above gnd.
Yo = 0.
T = 0.9 s. = Time in flight.
Dx = Xo*T = 27.5m/s * 0.9s = 24.75 m.
from launch point.
h = ho _ 0.5g*T^2,
h=11.2 - 4.9*(0.9)^2 =11.2-3.97=7.2 m.
above gnd.
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.