A speedboat moving at 28 m/s approaches a no-wake buoy marker 91 m ahead. The pilot slows the boat with a constant acceleration
1 answer:
Answer:
7.5 m/s
Explanation:
We can find its velocity when it reaches the buoy by applying one of Newton's equations of motion:
where v = final velocity
u = initial velocity
a = acceleration
s = distance traveled
From the question:
u = 28 m/s
a = -4
s = 91 m
Therefore:
The velocity of the boat when it reaches the buoy is 7.5 m/s.
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Answer:
The ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Explanation:
In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).
We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.
So, first we need to determine the components of the velocity of the ball, like this:
we pick the positive one, so it takes 0.317s for the ball to hit on point A.
so now we can find the distance from the net to point A with this time. We can find it like this:
Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:
t= -0.9159s or t=0.468s
we pick the positive one, so it takes 0.468s for the ball to hit on point B.
so now we can find the distance from the net to point B with this time. We can find it like this:
So once we got the two distances we can now find the difference between them:
so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.
