Question

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

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

Step-by-step solution

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.

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 \(v_{0x} = v_0 \cos(\theta)\)

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 \(v_{0y}\).

Calculate the vertical component of the initial speed

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

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.

Key concepts

Initial Speed Calculation

To understand how to calculate initial speed in projectile motion, it's essential to consider the factors involved in the movement. For a fish attempting to jump a waterfall, as in our example, the initial speed is influenced by the angle of the jump and the forces of gravity.

We begin by examining the vertical component since the height reached gives us tangible data about the jump. By using the vertical motion equation which includes gravitational acceleration, time, and height, we can solve for vertical speed. First, the formula used is:

• \[ h = v_{0y} t - \frac{1}{2}gt^2 \]

Where \(h\) is the waterfall height, \(v_{0y}\) is the initial vertical speed, \(t\) is time taken, and \(g\) is the acceleration due to gravity (9.8 m/s²). By rearranging to solve for \(v_{0y}\):

• \[ v_{0y} = \frac{1.05 + \frac{1}{2} \times 9.8 \times (2.1)^2}{2.1} \]

Once \(v_{0y}\) is calculated, we use it to find the initial speed (\(v_0\)) considering the jump angle:

• \[ v_0 = \frac{v_{0y}}{\sin(\theta)} \]

This formula acknowledges that the initial speed is affected by the launch angle, specifically its sine component.

Horizontal and Vertical Components

In projectile motion, breaking down the motion into horizontal and vertical components helps in understanding and calculating different elements like initial speed and time of flight. The components represent the direction of motion: horizontal (along the ground) and vertical (upwards or downwards).

The salmon's movement in our example demonstrates these components well. Using trigonometry, we decompose the total initial velocity (\(v_0\)) into horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)) parts. The equations involved are:

• \[ v_{0x} = v_0 \cos(\theta) \]
• \[ v_{0y} = v_0 \sin(\theta) \]

Where \(\theta\) is the launch angle. These expressions link the actual path of the salmon with its initial velocity conditions. By using these components, we could easily solve problems related to distance traveled horizontally or the maximum height achieved vertically.

Equations of Motion

The equations of motion are the backbone of solving projectile motion problems. They allow us to link the known variables (such as height, initial angle, and time) to unknowns like initial speed or distance traveled.

In our scenario, the salmon jumping the waterfall, we leveraged these standard equations. The relevant ones include:

• For vertical motion: \[ h = v_{0y} t - \frac{1}{2} g t^2 \]This equation describes how vertical motion is altered by the gravitational pull and the initial vertical velocity component.
• For horizontal motion: \[ x = v_{0x} t \]Describes how horizontal distance covered is dependent on time and the horizontal speed.

These equations clarify that both horizontal and vertical motions are happening simultaneously yet independently. This understanding is crucial for predicting and calculating different aspects of projectile trajectories. Mastering these formulas is key to solving any problem involving projectile motion, such as calculating the initial speed needed to jump a certain height or distance.