Problem 93 - Fundamentals of Physics

Question

A stone is thrown vertically upward. On its way up it passes point \(A\) with speed \(v\), and point \(B, 3.00 \mathrm{~m}\) higher than \(A,\) with speed \(\frac{1}{2} v .\) Calculate (a) the speed \(v\) and (b) the maximum height reached by the stone above point \(B\).

Step-by-Step Solution

Answer

The initial speed is 8.85 m/s, and the max height above B is 1.00 m.

Step 1: Understand the Concept

We need to calculate two things: the initial speed \(v\) at a lower point \(A\) and the maximum height achieved by the stone above point \(B\). The stone passes points \(A\) and \(B\) as it ascends, so we will use kinematic equations.

Step 2: Use Kinematics for Point B

The speed at point \(B\) is given as \(\frac{1}{2}v\) and point \(B\) is 3 meters above point \(A\). We use the kinematic equation \(v_B^2 = v_A^2 - 2g(y_B - y_A)\), where \(v_A = v\), \(v_B = \frac{1}{2}v\), and \(g = 9.8 \, \text{m/s}^2\).

Step 3: Establish the Equation

Substitute known values: \((\frac{1}{2}v)^2 = v^2 - 2 \times 9.8 \times 3\). Simplify: \(\frac{1}{4}v^2 = v^2 - 58.8\).

Step 4: Solve for Speed v

Rearrange the equation from Step 3: \(v^2 - \frac{1}{4}v^2 = 58.8\). Simplify to \(\frac{3}{4}v^2 = 58.8\). Solve for \(v^2\): \(v^2 = \frac{4}{3} \times 58.8 = 78.4\). So \(v = \sqrt{78.4} = 8.85 \, \text{m/s}\).

Step 5: Calculate Maximum Height Above B

To find the maximum height above \(B\), consider when the speed will reach zero. Use the kinematic equation with \(v = 0\) and initial speed at \(B\), \(v_B = 4.425 \, \text{m/s}\), as \(v^2 = v_B^2 - 2gh\).

Step 6: Solve for Maximum Height

Start with: \(0 = (4.425)^2 - 2 \times 9.8 \times h\). Rearrange to find \(h\): \(h = \frac{(4.425)^2}{2 \times 9.8} = \frac{19.58}{19.6} \approx 1.00 \, \text{m}\).

Key Concepts

Velocity Calculation

To find the initial velocity when dealing with projectile motion, particularly for objects thrown vertically, we use kinematic equations. The stone is thrown with a speed, decreases as it ascends, and increases once it descends due to gravity.
In the problem, the stone's speed when passing point A is given as \(v\). At point B, which is 3 meters higher, the stone’s speed reduces to half, or \(\frac{1}{2}v\).
To find the initial speed \(v\), we apply the kinematic equation:

• \(v_B^2 = v_A^2 - 2g(y_B - y_A)\)

We know

• \(v_B = \frac{1}{2}v\)
• \(g = 9.8 \, \text{m/s}^2\)
• \(y_B - y_A = 3\, \text{m}\)

Plug these into the equation to solve for \(v\). This involves substituting and rearranging, and you will find:

• \(\frac{1}{4}v^2 = v^2 - 58.8\)
• Rearrange to find \(\frac{3}{4}v^2 = 58.8\)
• \(v = \sqrt{78.4}\) which is approximately \(8.85 \, \text{m/s}\)

Maximum Height

The concept of maximum height in projectile motion is reached when the vertical speed becomes zero. After the stone passes point B with speed \(4.425 \, \text{m/s}\), it will continue to rise until its velocity is zero. This point marks the maximum height reached by the stone.
We use the kinematic equation to calculate the maximum height above point B.

• \(0 = v_B^2 - 2gh\)

We know

• \(v_B = 4.425 \, \text{m/s}\)
• \(g = 9.8 \, \text{m/s}^2\)

Rearranging the equation gives us:

• \(h = \frac{4.425^2}{2 \, \times \, 9.8}\)

Solving this shows that the height \(h\) above point B is approximately \(1.00 \, \text{m}\).
This distance is the additional height the stone will climb after passing point B before starting its descent.

Projectile Motion

Projectile motion involves the movement of an object under the influence of gravity, without considering air resistance. In vertical projectile motion, only vertical kinematics are evaluated.
The stone's motion is a classic example of one-dimensional vertical projectile motion, starting with an initial velocity against the force of gravity. Key concepts include:

• Objects thrown vertically will slow as they ascend and speed up as they descend due to gravity.
• The highest point, or apex, occurs when the vertical velocity becomes zero.

Key equations:

• The equation \(v = u + at\) isn't needed here as we work with given points' speeds.
• We've used \(v^2 = u^2 - 2g(s)\) to relate initial and final speeds and displacements.

Understanding these principles allows us to determine parameters like initial speed and maximum height, as shown in the previous sections.