Asked by Anonymous
A soccer ball is kicked with an initial horizontal velocity of 20 m/s and an initial vertical velocity of 16 m/s.
I've calculated:
Initial Speed: 25.61 m/s
Angle: 38.66 degrees
Maximum Height it reaches: 13.06 m
Distance kicked: 65.29m
NOW I NEED:
What is the speed of the ball 0.7 seconds after it was kicked?
How high above the ground is the ball 0.7 seconds after it is kicked?
help? please.
I've calculated:
Initial Speed: 25.61 m/s
Angle: 38.66 degrees
Maximum Height it reaches: 13.06 m
Distance kicked: 65.29m
NOW I NEED:
What is the speed of the ball 0.7 seconds after it was kicked?
How high above the ground is the ball 0.7 seconds after it is kicked?
help? please.
Answers
Answered by
Steve
the horizontal speed hs does not change
the vertical speed vs is 16-9.8t
vs(.7) = 16-9.8*.7 = 9.14
so, the speed is sqrt(9.14^2 + 20^2) = 21.99 m/s
vertical distance d = 1/2 at^2
d(.7) = 4.9*.7^2 = 2.4m
the vertical speed vs is 16-9.8t
vs(.7) = 16-9.8*.7 = 9.14
so, the speed is sqrt(9.14^2 + 20^2) = 21.99 m/s
vertical distance d = 1/2 at^2
d(.7) = 4.9*.7^2 = 2.4m
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.