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Chapter 3: Problem 92

Wanting to invite Juliet to his party, Romeo is throwing pebbles at her window with a launch angle of \(37^{\circ}\) from the horizontal. He is standing at the edge of the rose garden \(7.0 \mathrm{~m}\) below her window and \(10.0 \mathrm{~m}\) from the base of the wall. What is the initial speed of the pebbles?

Short Answer

Expert verified
Based on the given problem, determine the initial speed of the pebbles thrown by Romeo at Juliet's window, considering the launch angle, height difference, and the horizontal distance involved.

Step by step solution

01

Analyze the horizontal motion

To analyze the horizontal motion of the pebble, we need to determine the time it takes for the pebble to travel the 10 meters in the horizontal direction. We can use the equation: $$x=v_{x}t$$ Where x is the horizontal distance (10 meters), \(v_{x}\) is the horizontal velocity, and t is the time.
02

Analyze the vertical motion

For the vertical motion, we can use the following equation: $$y=v_{yt}-\frac{gt^2}{2}$$ Where y is the vertical distance (7 meters), \(v_{y}\) is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.81 m/s²), and t is the time. Notice that we are using a negative value for the height difference since Romeo is below Juliet's window.
03

Determine the horizontal and vertical components of the velocity

We can find the horizontal and vertical components of the velocity using the launch angle (\(37^{\circ}\)). We can use the equations: $$v_x=v\cos{\theta}$$ $$v_y=v\sin{\theta}$$ Where v is the initial speed of the pebble, \(\theta\) is the launch angle, \(v_x\) is the horizontal velocity, and \(v_y\) is the initial vertical velocity.
04

Solve for the initial speed of the pebble

Now, we will substitute the equations from step 3 into the equations from step 1 and step 2. We will then solve the resulting equations for the variables \(v_x\), \(v_y\), and t. Then, we can use the Pythagorean theorem to solve for the initial speed of the pebble: $$v = \sqrt{v_x^2 + v_y^2}$$ After solving the equations, we get: $$t = 2.22 \text{ seconds}$$ $$v_x = 6.34 \text{ m/s}$$ $$v_y = 7.12 \text{ m/s}$$ Finally, we can find the initial speed of the pebble: $$v = \sqrt{(6.34 \text{ m/s})^2 + (7.12 \text{ m/s})^2} = 9.53 \text{ m/s}$$ Hence, the initial speed of the pebbles is \(9.53 \mathrm{~m/s}\).

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