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Chapter 3: Problem 46

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

Expert verified
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

01

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: \(t = 1.40 \, \mathrm{s}\)
02

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: \(x = v_{0x} \cdot t = 30.0 \, \mathrm{m/s} \cdot 1.40 \, \mathrm{s} = 42.0 \, \mathrm{m}\)
03

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: \(y = 9.60 \, \mathrm{m} + 0 - \frac{1}{2}(9.81 \, \mathrm{m/s^2})(1.40 \, \mathrm{s})^2\) \(y = 9.60 \, \mathrm{m} - 9.676 \, \mathrm{m}\) \(y = -0.076 \, \mathrm{m}\)
04

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.

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Most popular questions from this chapter

You are serving as a consultant for the newest James Bond film. In one scene, Bond must fire a projectile from a cannon and hit the enemy headquarters located on the top of a cliff \(75.0 \mathrm{m}\) above and \(350 \mathrm{m}\) from the cannon. The cannon will shoot the projectile at an angle of \(40.0^{\circ}\) above the horizontal. The director wants to know what the speed of the projectile must be when it is fired from the cannon so that it will hit the enemy headquarters. What do you tell him? [Hint: Don't assume the projectile will hit the headquarters at the highest point of its flight. \(]\)
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