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Chapter 2: Problem 44

please assume the free-fall acceleration \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) unless a more precise value is given in the problem statement. Ignore air resistance. A brick is thrown vertically upward with an initial speed of $3.00 \mathrm{m} / \mathrm{s}\( from the roof of a building. If the building is \)78.4 \mathrm{m}$ tall, how much time passes before the brick lands on the ground?

Short Answer

Expert verified
The free-fall acceleration due to gravity is \(9.80\: m/s^2\). Answer: It takes approximately \(8.00\,s\) for the brick to land on the ground.

Step by step solution

01

Identify given information and note the equations needed

We are given the initial speed (\(v_0 = 3.00\: m/s\)), the height of the building (\(h = 78.4\: m\)), and the free-fall acceleration due to gravity (\(g = 9.80\: m/s^2\)). We will use the kinematic equation: \(h = v_0t - \dfrac{1}{2}gt^2\) to find the time it takes for the brick to land on the ground.
02

Solve the quadratic equation for time

Since the brick is thrown upward, the initial velocity is negative. The equation becomes: \(78.4 = -3.00t - \dfrac{1}{2}(9.80)t^2\) To solve this quadratic equation, we need to rewrite it in standard form: \(at^2 + bt + c = 0\) After rewriting, we get: \(4.9t^2 + 3.00t - 78.4 = 0\)
03

Solve for time t using the quadratic formula

Solve for time 't' using the quadratic formula: \(t = \dfrac{-b 卤 \sqrt{b^2 - 4ac}}{2a}\) Where \(a = 4.9\), \(b = 3.00\), and \(c = -78.4\) Plug in the values, and we get: \(t = \dfrac{-3.00 卤 \sqrt{(3.00)^2 - 4(4.9)(-78.4)}}{2(4.9)}\) Calculate the two possible values for t: \(t_1 = 2.0101\,s\) \(t_2 = 7.9899\,s\)
04

Choose the right solution for time

Since the brick lands on the ground, it will take longer than just reaching maximum height and returning to its initial position. Therefore, the correct time value is \(t_2 = 7.9899\,s\) (approximately \(8.00\,s\)).

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