Clarke Bolton
02/12/2023 · Junior High School

Question 4 (Mandatory) (1 point) A quarter is flipped up in the air at the start of a football game and returns to the point from which it was flipped after 2.5 s . Find the quarter's original speed. The acceleration of the quarter is \( 9.8 \mathrm{~m} / \mathrm{s} / \mathrm{s} \) downward. \( 4.9 \mathrm{~m} / \mathrm{s} \) \( 25 \mathrm{~m} / \mathrm{s} \) \( 12.25 \mathrm{~m} / \mathrm{s} \) \( 49 \mathrm{~m} / \mathrm{s} \)

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Given: - The quarter returns to the point from which it was flipped after 2.5 seconds. - The acceleration of the quarter is \(9.8 \, \text{m/s}^2\) downward. We can use the kinematic equation to find the original speed of the quarter. The kinematic equation for an object under constant acceleration is: \[ v = u + at \] where: - \( v \) is the final velocity, - \( u \) is the initial velocity, - \( a \) is the acceleration, and - \( t \) is the time. Since the quarter returns to the point from which it was flipped, the final velocity is 0. The acceleration is \(9.8 \, \text{m/s}^2\) downward, and the time is 2.5 seconds. Substitute the known values into the equation: \[ 0 = u + (-9.8) \times 2.5 \] Solve for \( u \) to find the original speed of the quarter. Solve the equation by following steps: - step0: Solve for \(u\): \(0=u+\left(-9.8\right)\times 2.5\) - step1: Simplify: \(0=u-24.5\) - step2: Swap the sides: \(u-24.5=0\) - step3: Move the constant to the right side: \(u=0+24.5\) - step4: Add the terms: \(u=24.5\) The original speed of the quarter is \( 24.5 \, \text{m/s} \) or \( 24\frac{1}{2} \, \text{m/s} \) or \( \frac{49}{2} \, \text{m/s} \). Therefore, the correct answer is \( 24.5 \, \text{m/s} \).

Quick Answer

The original speed of the quarter is \( 24.5 \, \text{m/s} \).
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