Jump Over 8 Cars: Calculating Minimum Speed for a Stunt Driver

What is the minimum speed that a stunt driver must drive off a horizontal ramp to jump over 8 cars parked side by side below? The minimum speed that a stunt driver must drive off a horizontal ramp to jump over 8 cars is 39.78 m/s.

Velocity is an essential concept in physics that measures the rate at which an object changes its position. In this scenario, a stunt driver aims to make his car jump over 8 cars parked below a horizontal ramp. The minimum speed required for this daring feat is crucial to ensure a successful jump.

To calculate the minimum speed, we consider the vertical height of the ramp, which is 1.5 meters, and the horizontal distance the driver must clear, which is 22 meters. By applying the second equation of motion and finding the time taken to cover the vertical distance, we can determine that the minimum speed needed is 39.78 m/s.

Velocity is the ratio of the distance traveled to the time taken, and in this case, the stunt driver must achieve a speed of 39.78 m/s to successfully jump over the 8 cars. This minimum speed ensures that the car clears the distance required for the stunt.

Therefore, with careful calculations and understanding of velocity, the stunt driver can achieve the necessary speed to make the thrilling jump over the parked cars. Safety precautions and precise measurements play a crucial role in such high-stakes maneuvers.

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