The water in a river flows uniformly at a constant speed of 2.53 m/s between parallel banks 69.8 m apart. You are to deliver a package directly across the river, but you can swim only at 1.74 m/s.
(a) If you choose to minimize the time you spend in the water, in what direction should you head? ____° from the direction of the stream
(b) How far downstream will you be carried? ______m
(c) If you choose to minimize the distance downstream that the river carries you, in what direction should you head? ____° from the direction of the stream
(d) How far downstream will you be carried? _____m
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I know we're dealing with motion in 2D, but I don't even know which equations to use or how to start. Any help or jumpstarts will be greatly appreciated.
A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 7.70 m/s at an angle of 23.0° below the horizontal. It strikes the ground 6.00 s later.
(a) How far horizontally from the base of the building does the ball strike the ground?
(b) Find the height from which the ball was thrown.
(c) How long does it take the ball to reach a point 10.0 m below the level of launching?
To solve this problem, we can break it down into two components - the horizontal and vertical components. Let's start with the first part:
(a) To minimize the time spent in the water, we need to find the direction in which you should head. Since you are swimming with a speed of 1.74 m/s and the river is flowing with a speed of 2.53 m/s, you want to swim in a direction that allows you to counteract the downstream drift caused by the river's flow.
To find the direction, we can use the concept of relative velocity. The relative velocity is the vector sum of your swimming speed and the river's flow velocity. We can represent these velocities as vectors, with your swimming velocity pointed directly across the river (perpendicular to its flow), and the river's velocity is parallel to the banks.
Now, let's consider the velocity vectors: Your swimming velocity is 1.74 m/s perpendicular to the river's current, and the river's velocity is 2.53 m/s parallel to the banks. To find the direction, we can imagine these vectors as the legs of a right triangle, and the resulting velocity vector (the relative velocity) is the hypotenuse.
Using the Pythagorean theorem, we can calculate the magnitude of the relative velocity:
relative velocity = sqrt(swimming velocity^2 + river velocity^2)
relative velocity = sqrt((1.74 m/s)^2 + (2.53 m/s)^2) = 3.07 m/s
Now that we have the magnitude of the relative velocity, we need to find the angle at which you should head. We can use trigonometry to do that:
cos(theta) = (river velocity) / (relative velocity)
cos(theta) = (2.53 m/s) / (3.07 m/s)
Taking the inverse cosine (arccos) of both sides, we can find theta:
theta = arccos((2.53 m/s) / (3.07 m/s))
Using a calculator, the angle theta is approximately 36.3 degrees.
So the direction you should head to minimize the time spent in the water is approximately 36.3 degrees from the direction of the stream.
(b) Now, let's calculate how far downstream you will be carried if you swim directly across the river. Since you are swimming perpendicular to the river's flow, the downstream distance can be calculated as follows:
Distance downstream = river velocity * time taken to cross the river
The time taken to cross the river can be found by dividing the distance across the river by your swimming velocity:
Time taken to cross the river = distance across the river / swimming velocity
Time taken to cross the river = 69.8 m / 1.74 m/s = 40.11 seconds (rounded to two decimal places)
Distance downstream = (2.53 m/s) * (40.11 seconds) = 101.53 meters
So, if you swim directly across the river, you will be carried approximately 101.53 meters downstream.
(c) To minimize the distance downstream that the river carries you, you need to swim at an angle that directly counteracts the river's flow. This means swimming directly against the river's flow.
(d) If you swim directly against the river's flow, you will not be carried downstream. Therefore, the distance downstream that you will be carried is zero.