Asked by jt
a contestant projects a coin with a speed of 7 m/s at an angleof 60 degrees to the horizontal. When the coin leaves his hand, the horizontal distance between the coin and the dish is 2.8m. The coin lands in the dish. calculate the horizontal component of the initial velocity of the coin?
Answers
Answered by
Damon
u = 7 cos 60 the whole time
so u = 7(1/2) = 3.5 m/s
end of problem
If there are more parts to this problem:
Vi = 7 sin 60 at the beginning
horizontal problem:
x = u t
2.8 = 7 (1/2) t
t = . 8 seconds in the air
t at top = .4 s
h = Vi (.4) - (9.81/2)(.4^2)
so u = 7(1/2) = 3.5 m/s
end of problem
If there are more parts to this problem:
Vi = 7 sin 60 at the beginning
horizontal problem:
x = u t
2.8 = 7 (1/2) t
t = . 8 seconds in the air
t at top = .4 s
h = Vi (.4) - (9.81/2)(.4^2)
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.