Question

An object is thrown upward with a speed of \(28.0 \mathrm{~m} / \mathrm{s}\). How long does it take it to reach its maximum height?

Short answer

Answer: The time it takes for the object to reach its maximum height is approximately 2.86 seconds.

Step-by-step solution

Identify the given values and the equation to be used

We are given the initial velocity of the object, \(v_0 = 28.0 \mathrm{~m} / \mathrm{s}\). We want to find the time \(t\) it takes to reach maximum height. At the maximum height, the object's final velocity \(v\) is 0 m/s. We can use the equation of motion that relates velocities and time, which is given by: \(v = v_0 - gt\) Where \(g\) is the acceleration due to gravity (\(9.81 \mathrm{~m} / \mathrm{s}^2\)).

Plug in the given values into the equation and solve for t

Now, we will substitute the given values into the equation: \(0 = 28.0 \mathrm{~m} / \mathrm{s} - 9.81 \mathrm{~m} / \mathrm{s}^2 \cdot t\) To solve for \(t\), we will first isolate the term with \(t\) on one side of the equation: \(9.81 \mathrm{~m} / \mathrm{s}^2 \cdot t = 28.0 \mathrm{~m} / \mathrm{s}\) Now, we will divide both sides of the equation by \(9.81 \mathrm{~m} / \mathrm{s}^2\) to find the value of \(t\): \(t = \frac{28.0 \mathrm{~m} / \mathrm{s}}{9.81 \mathrm{~m} / \mathrm{s}^2}\)

Calculate the time t and give the final answer

By dividing the numbers, we find the value of \(t\): \(t = \frac{28.0}{9.81} \approx 2.86 \mathrm{~s}\) It takes approximately \(2.86\) seconds for the object to reach its maximum height.