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Chapter 3: Problem 76

Salmon often jump upstream through waterfalls to reach their breeding grounds. One salmon came across a waterfall \(1.05 \mathrm{~m}\) in height, which she jumped in \(2.1 \mathrm{~s}\) at an angle of \(35^{\circ}\) to continue upstream. What was the initial speed of her jump?

Short Answer

Expert verified
Answer: The initial speed of the salmon's jump is approximately 7.90 m/s.

Step by step solution

01

Visualize and break down the problem

First, visualize the salmon's motion, which resembles projectile motion because it jumps at an angle to reach the top of the waterfall. We must break the problem down into horizontal and vertical components.
02

Write down the known values and equations

The height of the waterfall, \(h = 1.05 \mathrm{~m}\), and the time it took the salmon to jump, \(t = 2.1 \mathrm{~s}\). The angle of jump, \(\theta = 35^{\circ}\). We need to find the initial speed, \(v_0\). To do this, we will use the equations of motion: For the vertical component: \(h = v_{0y} t - \frac{1}{2}gt^2\) For horizontal component: \(x = v_{0x} t\) In both cases, \(v_{0y} = v_0 \sin(\theta)\) and \(v_{0x} = v_0 \cos(\theta)\)
03

Set up the equation for the vertical component

Using the vertical motion equation, plug in the known values and solve for \(v_{0y}\). \(h = v_{0y} t - \frac{1}{2}gt^2\) \(1.05 = v_{0y} \cdot 2.1 - \frac{1}{2} \cdot 9.8 \cdot (2.1)^2\) Solve this equation for \(v_{0y}\).
04

Calculate the vertical component of the initial speed

Rearrange the equation and solve for \(v_{0y}\): \(v_{0y} = \frac{1.05 + \frac{1}{2} \cdot 9.8 \cdot (2.1)^2}{2.1}\) \(v_{0y} \approx 4.57 \mathrm{~m/s}\)
05

Calculate the initial speed

Now, we know \(v_{0y}\) and need to find \(v_0\). We will use the fact that \(v_{0y} = v_0 \sin(\theta)\). Rearrange the equation and solve for \(v_0\). \(v_0 = \frac{v_{0y}}{\sin(\theta)}\) \(v_0 = \frac{4.57}{\sin(35^{\circ})}\) \(v_0 \approx 7.90 \mathrm{~m/s}\) The initial speed of the salmon's jump is approximately 7.90 m/s.

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Most popular questions from this chapter

A car drives straight off the edge of a cliff that is \(60.0 \mathrm{~m}\) high. The police at the scene of the accident note that the point of impact is \(150 . \mathrm{m}\) from the base of the cliff. How fast was the car traveling when it went over the cliff?

Some rental cars have a GPS unit installed, which allows the rental car company to check where you are at all times and thus also know your speed at any time. One of these rental cars is driven by an employee in the company's lot and, during the time interval from 0 to \(10 \mathrm{~s}\), is found to have a position vector as a function of time of $$ \begin{aligned} \vec{r}(t)=&\left((24.4 \mathrm{~m})-t(12.3 \mathrm{~m} / \mathrm{s})+t^{2}\left(2.43 \mathrm{~m} / \mathrm{s}^{2}\right)\right.\\\ &\left.(74.4 \mathrm{~m})+t^{2}\left(1.80 \mathrm{~m} / \mathrm{s}^{2}\right)-t^{3}\left(0.130 \mathrm{~m} / \mathrm{s}^{3}\right)\right) \end{aligned} $$ a) What is the distance of this car from the origin of the coordinate system at \(t=5.00 \mathrm{~s} ?\) b) What is the velocity vector as a function of time? c) What is the speed at \(t=5.00 \mathrm{~s} ?\) Extra credit: Can you produce a plot of the trajectory of the car in the \(x y\) -plane?

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A projectile is launched at a \(60^{\circ}\) angle above the horizontal on level ground. The change in its velocity between launch and just before landing is found to be \(\Delta \vec{v}=\vec{v}_{\text {landing }}-\vec{v}_{\text {launch }}=-20 \hat{y} \mathrm{~m} / \mathrm{s}\). What is the initial velocity of the projectile? What is its final velocity just before landing?
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