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Chapter 2: Problem 61

A ball is thrown directly downward, with an initial speed of \(10.0 \mathrm{~m} / \mathrm{s}\), from a height of \(50.0 \mathrm{~m}\). After what time interval does the ball strike the ground?

Short Answer

Expert verified
Answer: The ball strikes the ground after about 2.074 seconds.

Step by step solution

01

Write the given values and the equation of motion

We are given the following values: - Initial velocity: \(v_{0y} = 10.0 \, \mathrm{m/s}\) - Height: \(y = 50.0 \, \mathrm{m}\) - Acceleration due to gravity: \(a = 9.81 \, \mathrm{m/s^2}\) The equation of motion we will use is: \(y = v_{0y}t + \frac{1}{2}at^2\)
02

Plug in the given values and solve for time 't'

We can now plug in the given values into the equation: \(50.0 = 10.0t + \frac{1}{2}(9.81)t^2\) To solve this equation for the time interval \(t\), we can first rearrange the equation to form a quadratic equation. \(0 = \frac{1}{2}(9.81)t^2 + 10.0t - 50.0\) This forms a quadratic equation of the form \(0 = at^2 + bt + c\), where \(a = \frac{1}{2}(9.81)\), \(b = 10.0\), and \(c = -50.0\). We can solve this equation for \(t\) using the quadratic formula, which is given by: \(t = \cfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
03

Calculate the time interval 't'

We can now compute the time interval \(t\) using the quadratic formula: \(t = \cfrac{-10.0 \pm \sqrt{(10.0)^2 - 4(\frac{1}{2}(9.81))(-50.0)}}{2(\frac{1}{2}(9.81))}\) By solving the above expression, we will get two values for time. However, since time cannot be negative, we will choose the positive value. \(t \approx 2.074 \, \mathrm{s}\) The ball strikes the ground after about \(2.074 \, \mathrm{s}\).

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Most popular questions from this chapter

The fastest speed in NASCAR racing history was \(212.809 \mathrm{mph}\) (reached by Bill Elliott in 1987 at Talladega). If the race car decelerated from that speed at a rate of \(8.0 \mathrm{~m} / \mathrm{s}^{2},\) how far would it travel before coming to a stop?

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