Chapter 2: Problem 62

An object is thrown vertically upward and has a speed of \(20.0 \mathrm{~m} / \mathrm{s}\) when it reaches two thirds of its maximum height above the launch point. Determine its maximum height.

Short Answer

Expert verified

Answer: The maximum height reached by the object is approximately 40.61 m.

Step by step solution

Identify the known variables

We know the following: - Speed at two thirds of maximum height, \(v = 20.0\mathrm{~m/s}\) - Fraction of the maximum height, \(h' = \dfrac{2}{3} H\) - Acceleration due to gravity, \(a = -9.81\mathrm{~m/s^2}\) (downward, hence negative) - Final velocity at maximum height, \(v_f = 0\mathrm{~m/s^2}\) Our goal is to find the maximum height, \(H\).

Use the equation of motion to find the initial velocity at two thirds height

The equation of motion that relates the velocity, the initial velocity, the height, and the acceleration is: \(v^2 = v_i^2 + 2a(h - h_i)\). At two thirds of the maximum height, the above equation becomes: \(v^2 = v_i^2 + 2a(h' - h_i)\). Since the object is thrown from the ground, we know that \(h_i = 0\). So, we have: \(v^2 = v_i^2 + 2a h'\). Solve for \(v_i\) using the given values: \((20.0\mathrm{~m/s})^2 = v_i^2 + 2(-9.81\mathrm{~m/s^2})(\dfrac{2}{3}H)\).

Determine initial velocity at the ground

Rearrange the equation above to find the initial velocity: \(v_i^2 = (20.0\mathrm{~m/s})^2 - 2(-9.81\mathrm{~m/s^2})(\dfrac{2}{3}H)\).

Calculate the final velocity at maximum height

Since the object reaches the maximum height and momentarily comes to rest, we know its final velocity, \(v_f = 0\mathrm{~m/s}\). Now we can use another equation of motion to relate the initial and final velocities, height, and acceleration: \(v_f^2 = v_i^2 + 2aH\).

Plug in the initial velocity equation to find the maximum height

Substitute the equation for \(v_i^2\) found in step 3 into the equation of motion in step 4: \(0\mathrm{~m/s^2} = ((20.0\mathrm{~m/s})^2 - 2(-9.81\mathrm{~m/s^2})(\dfrac{2}{3}H))+ 2(-9.81\mathrm{~m/s^2})H\).

Solve for the maximum height H

Now we can solve for \(H\). First, simplify the equation: \(0 = (20.0\mathrm{~m/s})^2 - (\dfrac{4}{3})(-9.81\mathrm{~m/s^2})H + 2(-9.81\mathrm{~m/s^2})H\). Next, rearrange and solve for \(H\): \(H = \dfrac{(20.0\mathrm{~m/s})^2}{(\dfrac{-2}{3})(-9.81\mathrm{~m/s^2})}\). \(H \approx 40.61\mathrm{~m}\). The maximum height reached by the object is approximately \(40.61\mathrm{~m}\).

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An object is thrown vertically upward and has a speed of \(20.0 \mathrm{~m} / \mathrm{s}\) when it reaches two thirds of its maximum height above the launch point. Determine its maximum height.

Short Answer

Step by step solution

Identify the known variables

Use the equation of motion to find the initial velocity at two thirds height

Determine initial velocity at the ground

Calculate the final velocity at maximum height

Plug in the initial velocity equation to find the maximum height

Solve for the maximum height H

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