A ball is released at the top of a ramp at t =0. which is the speed of the ball at t=4
2 answers:
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Answer:
a)
b) Ball 1 has a greater speed than ball 2 when they are passing.
c) The height is the same for both balls = 3h/4.
Explanation:
a) We can find the time when the two balls meet by equating the distances as follows:
Where:
: is the initial height = h
: is the initial speed of ball 1 = 0 (it is dropped from rest)
(1)
Now, for ball 2 we have:
Where:
: is the initial height of ball 2 = 0
(2)
By equating equation (1) and (2) we have:
Where the initial velocity of the ball 2 is:
Since = 0 at the maximum height (h):
Hence, the time when they pass each other is:
b) When they are passing the speed of each one is:
For ball 1:
The minus sign is because ball 1 is going down.
For ball 2:
Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.
c) The height of the ball is:
For ball 1:
For ball 2:
Then, when they are passing the height is the same for both = 3h/4.
I hope it helps you!