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Chapter 3: Problem 37

Jack and Jill ran up the hill at \(3.0 \mathrm{m} / \mathrm{s}\). The horizontal component of Jill's velocity vector was \(2.5 \mathrm{m} / \mathrm{s}\) a. What was the angle of the hill? b. What was the vertical component of Jill's velocity?

Short Answer

Expert verified
a. The angle of the hill is \(\theta = \cos^{-1}(0.83333333333)\) degrees. b. The vertical component of Jill's velocity is 1.5 m/s.

Step by step solution

01

Find the angle of the hill

Jill's velocity up the hill forms the hypotenuse of a right triangle, while the horizontal component of her velocity is one of the legs of the triangle. The angle \(\theta\) can be found using the relation: \n\n\t\(\cos\theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}}\). \n\nHere, the adjacent side is the horizontal component of velocity (2.5 m/s) and the hypotenuse is the magnitude of velocity (3.0 m/s). Applying these values, we get: \n\n\t\(\cos\theta = \frac{2.5 \mathrm{m/s}}{3.0 \mathrm{m/s}}\).\n\nHence, the angle can be obtained by taking the cosine inverse of the above equation.
02

Calculate the angle

Firstly according to the formula in previous step, calculate \(\cos\theta\). Then, find the angle via inverse cosine. So, \n\n\t\(\cos\theta = 0.83333333333\) \n\n\tand \n\n\t\(\theta = \cos^{-1}(0.83333333333)\).
03

Find the vertical component of Jill's velocity

Having obtained the angle, we can now proceed to the second part of the problem. The vertical component of velocity can be obtained by using Pythagoras theorem, which is: \n\n\t\(c^2 = a^2 + b^2\),\n\nwhere 'c' is the hypotenuse, and 'a' and 'b' are the other two sides of the right triangle. Here, 'c' is the magnitude of velocity (3.0 m/s), 'a' is the horizontal component of velocity (2.5 m/s), and 'b' is the vertical component of velocity (which we have to find). Let鈥檚 denote the vertical component of velocity as Vv. Then we get : \n\n\t\(Vv = \sqrt{c^2 - a^2}\).\n\nWhen we substitute the given values, we get: \n\n\t\(Vv = \sqrt{(3.0 \mathrm{m/s})^2 - (2.5 \mathrm{m/s})^2}\).\n\n
04

Calculate the vertical component of velocity

From the formula in the previous step, calculate the vertical component of velocity, which comes to: \n\n\t\(Vv = \sqrt{(3.0 \mathrm{m/s})^2 - (2.5 \mathrm{m/s})^2} = 1.5 \mathrm{m/s}\).

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