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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

Neglect air resistance for the following. A soccer ball is kicked from the ground into the air. When the ball is at a height of \(12.5 \mathrm{~m},\) its velocity is \((5.6 \hat{x}+4.1 \hat{y}) \mathrm{m} / \mathrm{s}\). a) To what maximum height will the ball rise? b) What horizontal distance will be traveled by the ball? c) With what velocity (magnitude and direction) will it hit the ground?

To determine the gravitational acceleration at the surface of a newly discovered planet, scientists perform a projectile motion experiment. They launch a small model rocket at an initial speed of \(50.0 \mathrm{~m} / \mathrm{s}\) and an angle of \(30.0^{\circ}\) above the horizontal and measure the (horizontal) range on flat ground to be \(2165 \mathrm{~m}\). Determine the value of \(g\) for the planet.

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