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The Incredible Hulk rolls a bowling ball with his left hand with an initial speed of 106.6 m/s. What is the maximum distance (in meters) the bowling ball will reach if it is accelerating at -6.3 m/s^2? Round to 1 decimal place.

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Final answer:

Using the kinematic equations of motion, the maximum distance reached by a bowling ball thrown by the Incredible Hulk with an initial velocity of 106.6 m/s and decelerating at -6.3 m/s² is approximately 901.7 meters.

Step-by-step explanation:

The question involves calculating the maximum distance a bowling ball would travel when thrown with an initial speed and subject to a decelerating force. To find the distance, we'll use the kinematic equation for motion under constant acceleration, which is s = vt - 0.5at², where s is the distance, v is the initial velocity, a is the acceleration, and t is time.

Since the Incredible Hulk does the throwing, his strength imparts an initial velocity of 106.6 m/s to the bowling ball. The ball then decelerates at a = -6.3 m/s² until it stops. We need to first calculate the time it takes for the bowling ball to stop by using the equation v = u +at where v is the final velocity (which is 0 when the ball stops), u is the initial velocity, and a is the acceleration. After finding the time, we will plug it into the distance equation to calculate the maximum distance traveled by the ball.

Following the calculation:

  • The time it takes for the bowling ball to come to a stop:
    t = -v / a = -106.6 m/s / (-6.3 m/s²) = 16.92 s
  • Now, using the time to find the distance:
    s = vt - 0.5at²
    s = 106.6 m/s * 16.92 s - 0.5 * (-6.3 m/s²) * (16.92 s)²
    s = 1802.392 - (0.5 * 6.3 * 286.1664)
    s = 1802.392 - 900.3172
    s = 901.67 m

Thus, the maximum distance the bowling ball will reach if it is accelerating at -6.3 m/s² is approximately 901.7 meters, rounding to one decimal place.

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