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Chapter 3: Problem 75

At the end of the spring term, a high school physics class celebrates by shooting a bundle of exam papers into the town landfill with a homemade catapult. They aim for a point that is \(30.0 \mathrm{~m}\) away and at the same height from which the catapult releases the bundle. The initial horizontal velocity component is \(3.90 \mathrm{~m} / \mathrm{s}\). What is the initial velocity component in the vertical direction? What is the launch angle?

Short Answer

Expert verified
Answer: The initial vertical velocity is 37.7 m/s, and the launch angle is approximately 84.2°.

Step by step solution

01

Calculate time of flight

Using the first equation: \(x = x_0 + v_{0x}t_{\text{flight}}\) Since the initial position \(x_0 = 0\), we can write: \(t_{\text{flight}} = \frac{x}{v_{0x}}\) Plugging in the given values, we get: \(t_{\text{flight}} = \frac{30.0 \mathrm{~m}}{3.90 \mathrm{~m} / \mathrm{s}} \approx 7.69 \mathrm{~s}\)
02

Calculate initial vertical velocity

Using the second equation: \(0 = v_{0y}t_{\text{flight}} - \frac{1}{2}gt^2_{\text{flight}}\) Solving for \(v_{0y}\), we get: \(v_{0y} = \frac{1}{2}gt_{\text{flight}}\) Using \(g = 9.81 \mathrm{~m} / \mathrm{s}^2\) and \(t_{\text{flight}} \approx 7.69 \mathrm{~s}\), we calculate \(v_{0y}\): \(v_{0y} = \frac{1}{2}(9.81 \mathrm{~m} / \mathrm{s}^2)(7.69 \mathrm{~s}) \approx 37.7 \mathrm{~m} / \mathrm{s}\)
03

Calculate launch angle

Using the third equations, we get \(\tan{\theta} = \frac{v_{0y}}{v_{0x}}\) Plugging in the known values: \(\tan{\theta} = \frac{37.7 \mathrm{~m} / \mathrm{s}}{3.90 \mathrm{~m} / \mathrm{s}} \approx 9.67\) Now, compute the launch angle: \(\theta = \arctan{(9.67)} \approx 84.2^\circ\) So, the initial vertical velocity component is \(37.7\mathrm{~m} /\mathrm{s}\), and the launch angle is approximately \(84.2^\circ\).

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