Questions: When you throw a pebble straight up with initial speed V, it reaches a maximum height H with no air resistance. At what speed should you throw it up vertically so it will go twice as high? A) 4 V B) 8 V C) sqrt(2) V D) 2 V E) 16 V

When you throw a pebble straight up with initial speed V, it reaches a maximum height H with no air resistance. At what speed should you throw it up vertically so it will go twice as high? A) 4 V B) 8 V C) sqrt(2) V D) 2 V E) 16 V
Transcript text: 14) When you throw a pebble straight up with initial speed $V$, it reaches a maximum height H with no air resistance. At what speed should you throw it up vertically so it will go twice as high? A) 4 V B) 8 V C) $\sqrt{2} \mathrm{~V}$ D) 2 V E) 16 V
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Solution

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Solution Steps

Step 1: Understand the Relationship Between Initial Speed and Maximum Height

When a pebble is thrown straight up with an initial speed \( V \), it reaches a maximum height \( H \). The relationship between the initial speed and the maximum height can be derived from the kinematic equation for vertical motion without air resistance:

\[ v^2 = u^2 + 2a s \]

where:

  • \( v \) is the final velocity (0 m/s at the maximum height),
  • \( u \) is the initial velocity (\( V \)),
  • \( a \) is the acceleration due to gravity (\(-g\)),
  • \( s \) is the displacement (maximum height \( H \)).

Substituting the known values, we get:

\[ 0 = V^2 - 2gH \]

Solving for \( H \), we have:

\[ H = \frac{V^2}{2g} \]

Step 2: Determine the New Height and Required Initial Speed

To find the speed required to reach twice the height, \( 2H \), we use the same kinematic equation:

\[ 0 = u'^2 - 2g(2H) \]

Solving for the new initial speed \( u' \):

\[ u'^2 = 4gH \]

Step 3: Relate the New Initial Speed to the Original Speed

Substitute the expression for \( H \) from Step 1 into the equation for \( u'^2 \):

\[ u'^2 = 4g \left(\frac{V^2}{2g}\right) \]

Simplifying:

\[ u'^2 = 2V^2 \]

Taking the square root of both sides:

\[ u' = \sqrt{2}V \]

Final Answer

The speed required to throw the pebble so it will go twice as high is \(\sqrt{2}V\).

\[ \boxed{\sqrt{2}V} \]

Thus, the correct answer is C) \(\sqrt{2} \mathrm{~V}\).

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