Questions: In the absence of air resistance two balls are thrown upward from the same launch point. Ball A rises to a maximum height above the launch point that is four times greater than that of ball B. The launch speed of ball A is times greater than that of ball B.

In the absence of air resistance two balls are thrown upward from the same launch point. Ball A rises to a maximum height above the launch point that is four times greater than that of ball B. The launch speed of ball A is times greater than that of ball B.

Transcript text: Question 13 of 17 View Policies Current Attempt in Progress In the absence of air resistance two balls are thrown upward from the same launch point. Ball A rises to a maximum height above the launch point that is four times greater than that of ball B. The launch speed of ball $A$ is $\qquad$ times greater than that of ball B. four two sixteen eight eTextbook and Media Save for Later Attempts: 0 of 10 used Submit Answer

Solution

Solution Steps

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

In the absence of air resistance, the maximum height $ h $ reached by a projectile is related to its initial launch speed $ v_0 $ by the equation:

\[ h = \frac{v_0^2}{2g} \]

where $ g $ is the acceleration due to gravity.

Step 2: Set Up the Equation for Both Balls

Let $ h_A $ and $ h_B $ be the maximum heights reached by balls A and B, respectively, and $ v_{0A} $ and $ v_{0B} $ be their initial launch speeds. According to the problem, $ h_A = 4h_B $.

Using the height equation for both balls, we have:

\[ h_A = \frac{v_{0A}^2}{2g} \]

\[ h_B = \frac{v_{0B}^2}{2g} \]

Step 3: Relate the Heights and Solve for the Speed Ratio

Since $ h_A = 4h_B $, we can substitute the expressions for $ h_A $ and $ h_B $:

\[ \frac{v_{0A}^2}{2g} = 4 \left(\frac{v_{0B}^2}{2g}\right) \]

Simplifying, we get:

\[ v_{0A}^2 = 4v_{0B}^2 \]

Taking the square root of both sides gives:

\[ v_{0A} = 2v_{0B} \]