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)
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?
1 answer
1 answer