Question

A baseball is thrown horizontally from a height of \(9.60 \mathrm{m}\) above the ground with a speed of \(30.0 \mathrm{m} / \mathrm{s}\) Where is the ball after 1.40 s has elapsed?

Short answer

Answer: The position of the baseball after 1.40 seconds is (42.0 m, -0.076 m). Note that the negative value of the vertical position indicates that the baseball has hit the ground before 1.40 seconds elapsed.

Step-by-step solution

Identify the given data

The initial height: \(h_0 = 9.60 \, \mathrm{m}\) The horizontal speed: \(v_{0x} = 30.0 \, \mathrm{m/s}\) The time elapsed: \(t = 1.40 \, \mathrm{s}\)

Analyze the horizontal motion

Since there is no horizontal acceleration, the horizontal motion is uniform. Therefore, we can use the equation \(x = v_{0x} t\) to find the horizontal position of the baseball: \(x = v_{0x} \cdot t = 30.0 \, \mathrm{m/s} \cdot 1.40 \, \mathrm{s} = 42.0 \, \mathrm{m}\)

Analyze the vertical motion

Since gravity is the only force acting on the baseball, we have a constant downward acceleration \(a_y = -9.81 \, \mathrm{m/s^2}\). We can now use the following kinematic equation to find the vertical position of the baseball at time \(t\): \(y = h_0 + v_{0y}t + \frac{1}{2}a_yt^2\) Here, \(v_{0y} = 0\) because the baseball is thrown horizontally. So, substituting the given values, we get: \(y = 9.60 \, \mathrm{m} + 0 - \frac{1}{2}(9.81 \, \mathrm{m/s^2})(1.40 \, \mathrm{s})^2\) \(y = 9.60 \, \mathrm{m} - 9.676 \, \mathrm{m}\) \(y = -0.076 \, \mathrm{m}\)

Find the position of the baseball at time t

The final position of the baseball after 1.40 seconds is given by the coordinates \((x, y)\). So, the position of the baseball is: \((42.0 \, \mathrm{m}, -0.076 \, \mathrm{m})\) Note that the negative value of the vertical position indicates that the baseball has hit the ground before 1.40 seconds elapsed.