Problem 5. A bullet leaves a rifle with a velocity of 452m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.93 m. Determine the acceleration of the bullet. square Problem 6. The observation deck of the Empire State Building is 381 m above the street.Determine the time required for a penny to free fall from the deck to the street below. square Problem 7. A bowling ball is dropped on the Jupiter from a height of 3 meters with an initial velocity of 0.4m/s The acceleration of gravity on Jupiter is 24.5m/s^2 How long does it take for the bowling ball to reach the surface of Jupiter? square Problem 8. A truck accelerates uniformly from rest to a speed of 9.2m/s over a distance of 26.2 m. Determine the acceleration of the truck. square

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Problem 5. A bullet leaves a rifle with a velocity of 452m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.93 m. Determine the acceleration of the bullet. square Problem 6. The observation deck of the Empire State Building is 381 m above the street.Determine the time required for a penny to free fall from the deck to the street below. square Problem 7. A bowling ball is dropped on the Jupiter from a height of 3 meters with an initial velocity of 0.4m/s The acceleration of gravity on Jupiter is 24.5m/s^2 How long does it take for the bowling ball to reach the surface of Jupiter? square Problem 8. A truck accelerates uniformly from rest to a speed of 9.2m/s over a distance of 26.2 m. Determine the acceleration of the truck. square

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# Brief Explanations: To solve these problems, we will use kinematic equations of motion. These equations relate the initial velocity, final velocity, acceleration, distance, and time. # Answer: **Problem 5:** Using the kinematic equation \( v^2 = u^2 + 2as \): - \( v = 452 \, \text{m/s} \) - \( u = 0 \, \text{m/s} \) (assuming the bullet starts from rest) - \( s = 0.93 \, \text{m} \) \[ 452^2 = 0 + 2 \cdot a \cdot 0.93 \] \[ a = \frac{452^2}{2 \cdot 0.93} \] \[ a \approx 109,934 \, \text{m/s}^2 \] # Answer: The acceleration of the bullet is approximately \( 109,934 \, \text{m/s}^2 \). **Problem 6:** Using the kinematic equation \( s = \frac{1}{2}gt^2 \): - \( s = 381 \, \text{m} \) - \( g = 9.8 \, \text{m/s}^2 \) \[ 381 = \frac{1}{2} \cdot 9.8 \cdot t^2 \] \[ t^2 = \frac{381 \cdot 2}{9.8} \] \[ t^2 \approx 77.76 \] \[ t \approx \sqrt{77.76} \] \[ t \approx 8.82 \, \text{seconds} \] # Answer: The time required for a penny to free fall from the deck to the street below is approximately \( 8.82 \, \text{seconds} \). **Problem 7:** Using the kinematic equation \( s = ut + \frac{1}{2}at^2 \): - \( s = 3 \, \text{m} \) - \( u = 0.4 \, \text{m/s} \) - \( a = 24.5 \, \text{m/s}^2 \) \[ 3 = 0.4t + \frac{1}{2} \cdot 24.5 \cdot t^2 \] \[ 3 = 0.4t + 12.25t^2 \] \[ 12.25t^2 + 0.4t - 3 = 0 \] Solving this quadratic equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ a = 12.25, \, b = 0.4, \, c = -3 \] \[ t = \frac{-0.4 \pm \sqrt{0.4^2 - 4 \cdot 12.25 \cdot (-3)}}{2 \cdot 12.25} \] \[ t = \frac{-0.4 \pm \sqrt{0.16 + 147}}{24.5} \] \[ t = \frac{-0.4 \pm \sqrt{147.16}}{24.5} \] \[ t = \frac{-0.4 \pm 12.13}{24.5} \] Taking the positive root: \[ t \approx \frac{11.73}{24.5} \] \[ t \approx 0.48 \, \text{seconds} \] # Answer: The time it takes for the bowling ball to reach the surface of Jupiter is approximately \( 0.48 \, \text{seconds} \). **Problem 8:** Using the kinematic equation \( v^2 = u^2 + 2as \): - \( v = 9.2 \, \text{m/s} \) - \( u = 0 \, \text{m/s} \) (starting from rest) - \( s = 26.2 \, \text{m} \) \[ 9.2^2 = 0 + 2 \cdot a \cdot 26.2 \] \[ a = \frac{9.2^2}{2 \cdot 26.2} \] \[ a \approx 1.62 \, \text{m/s}^2 \] # Answer: The acceleration of the truck is approximately \( 1.62 \, \text{m/s}^2 \).

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