Problem 47 - College Physics

Question

A ball is projected horizontally with an initial speed of \(5.0 \mathrm{~m} / \mathrm{s}\). Find its (a) position and (b) velocity at \(t=2.5 \mathrm{~s}\)

Step-by-Step Solution

Answer

Position: (12.5 m, -30.625 m), Velocity: ~25.0 m/s at an angle.

Step 1: Understand the problem

We have a ball that is projected horizontally with an initial speed of \(5.0 \, \text{m/s}\). We need to determine the ball's position and velocity after \(2.5\) seconds. The ball undergoes motion in two dimensions—horizontal and vertical. The horizontal motion is uniform, as there is no acceleration, while the vertical motion is influenced by gravity, causing constant acceleration.

Step 2: Analyze the Horizontal Motion

For horizontal motion, the velocity \(v_x\) is constant since there is no acceleration. Therefore, after time \(t\), the horizontal position \(x\) is given by:\[ x = v_x \cdot t = 5.0 \, \text{m/s} \times 2.5 \, \text{s} \]Calculate \(x\).

Step 3: Calculate Horizontal Position

Substituting the values, we have:\[ x = 5.0 \, \text{m/s} \times 2.5 \, \text{s} = 12.5 \, \text{m} \] Thus, the horizontal position at \(t = 2.5 \, \text{s}\) is \(12.5 \, \text{m}\).

Step 4: Analyze Vertical Motion

For vertical motion, the initial vertical velocity \(v_{y0}\) is \(0\), and the vertical acceleration is due to gravity, which is \(9.8 \, \text{m/s}^2\). The vertical position \(y\) is given by:\[ y = v_{y0} \cdot t + \frac{1}{2} g t^2 = 0 + \frac{1}{2} \cdot 9.8 \, \text{m/s}^2 \cdot (2.5 \, \text{s})^2 \]

Step 5: Calculate Vertical Position

Substitute the values:\[ y = \frac{1}{2} \cdot 9.8 \, \text{m/s}^2 \cdot 6.25 \, \text{s}^2 = \frac{1}{2} \cdot 9.8 \, \text{m/s}^2 \cdot 6.25 = 30.625 \, \text{m} \]Thus, the vertical position at \(t = 2.5 \, \text{s}\) is \(-30.625 \, \text{m}\) (negative because the ball is falling downwards).

Step 6: Determine the Ball's Velocity

The horizontal velocity \(v_x\) remains \(5.0 \, \text{m/s}\). The vertical velocity \(v_y\) at time \(t\) is given by:\[ v_y = v_{y0} + g \, t = 0 + 9.8 \, \text{m/s}^2 \cdot 2.5 \, \text{s} \]

Step 7: Calculate Vertical Velocity

Substituting the values, we have:\[ v_y = 9.8 \, \text{m/s}^2 \cdot 2.5 \, \text{s} = 24.5 \, \text{m/s} \]The vertical velocity is \(24.5 \, \text{m/s}\) downwards.

Step 8: Find Resultant Velocity

The velocity of the ball is the vector sum of its horizontal and vertical components. Use the Pythagorean theorem to find the magnitude:\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(5.0 \, \text{m/s})^2 + (24.5 \, \text{m/s})^2} \]

Step 9: Calculate Resultant Velocity

Calculate the magnitude:\[ v = \sqrt{25 + 600.25} = \sqrt{625.25} \approx 25.0 \, \text{m/s} \]Therefore, the magnitude of the velocity at \(t = 2.5 \, \text{s}\) is approximately \(25.0 \, \text{m/s}\).

Step 10: Summarize the Results

At \(t = 2.5 \, \text{s}\), the ball's position is \((12.5 \, \text{m}, -30.625 \, \text{m})\), and its velocity is approximately \(25.0 \, \text{m/s}\) at an angle below the horizontal.

Key Concepts

Horizontal Motion

When we talk about horizontal motion, we're discussing how objects move across a flat surface without any upward or downward changes in their path. This kind of motion is interesting because it happens at a constant speed, meaning the object doesn’t speed up or slow down. In the case of the ball from our exercise, it was launched with a speed of \(5.0 \, \text{m/s}\). Hence, its horizontal speed remained steady throughout the motion.

This constant speed is due to the absence of horizontal forces like friction or air resistance (we're assuming these are negligible here). Thus, the horizontal distance covered by the ball in a certain amount of time can be determined using the simple equation:

• \(x = v_x \times t\)

For this problem, we have \(v_x = 5.0 \, \text{m/s}\) and \(t = 2.5 \, \text{s}\), and when we plug these into our equation, we find the horizontal position becomes \(12.5 \, \text{m}\). This calculation shows the predictable path objects follow when moving straight without changing speed.

Vertical Motion

Vertical motion involves the movement of an object in a direction perpendicular to the horizontal plane, usually influenced by gravity. It’s different from horizontal motion because here, the object accelerates as it moves downward. The primary influence is the gravitational force, which gives a constant acceleration of approximately \(9.8 \, \text{m/s}^2\).

In the exercise, the ball starts with no initial vertical speed \(v_{y0} = 0\). Over time, gravity pulls the ball downward, and so it accelerates in the vertical direction. We use the following formula to determine the vertical position:

• \( y = v_{y0} \cdot t + \frac{1}{2} g t^2\)

Substituting the values for gravity \(g\), time \(t = 2.5 \, \text{s}\), and knowing the initial velocity is zero, we find that \( y = -30.625 \, \text{m}\). The negative sign indicates it has fallen below its starting point, showing how gravity consistently increases an object's speed downward.

Constant Acceleration

Acceleration describes how quickly the velocity of an object changes over time. In projectile motion, most notably in vertical motion, we deal with a constant acceleration due to gravity, \(9.8 \, \text{m/s}^2\). This consistent force causes the speed of an object to increase steadily as it falls.

For our exercise, we calculate the vertical component of the ball's velocity using the following relation:

• \(v_y = v_{y0} + g \times t\)

Given that the ball starts with no vertical velocity \(v_{y0} = 0\), after \(2.5 \, \text{s}\), it attains a velocity of \(24.5 \, \text{m/s}\) downward (as per gravity's pull). This illustrates how constant acceleration acts over time, increasing the velocity linearly as long as the force (gravity in this case) remains unchanged. Understanding constant acceleration is crucial because it helps predict how objects will move under the influence of steady forces.