Minimum Speed for Stunt Driver to Safely Jump Over Cars

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SUMMARY

The minimum speed required for a stunt driver to safely jump over ten cars from a height of 1.25 meters and clear a distance of 20.0 meters is determined by analyzing the time taken to fall and the horizontal distance traveled. The driver must first calculate the time to fall 1.25 meters using the equation of motion, then use that time to find the necessary horizontal velocity to cover 20.0 meters. This involves setting up two equations: one for vertical motion and one for horizontal motion, ensuring the calculations account for the height difference and horizontal distance.

PREREQUISITES
  • Understanding of basic physics concepts, specifically projectile motion.
  • Familiarity with kinematic equations for vertical and horizontal motion.
  • Ability to solve equations involving time, distance, and velocity.
  • Knowledge of vector components in motion analysis.
NEXT STEPS
  • Study the kinematic equations for projectile motion in detail.
  • Learn how to calculate time of flight for objects in free fall.
  • Explore vector decomposition in physics for better understanding of motion.
  • Practice solving similar problems involving horizontal jumps and vertical drops.
USEFUL FOR

Physics students, stunt coordinators, and anyone interested in understanding the mechanics of jumps and projectile motion in practical scenarios.

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Homework Statement


A stunt driver sets a jump off a horizontal platform and over ten cars. The height of the platform is 1.25 m, and the total distance of the cars is 20.0 m. Find the minimum speed required to safely clear the cars.


Homework Equations


x - x initial = velocity X t + 1/2 a t squared (??)


The Attempt at a Solution


I tried to follow an example from the book but got confused because of the height difference (the book example did not have a change in y). I looked at solving for various vector components, but I was unsure of the angle...if the driver is going off a flat surface, would the angle be 0? I figured that the change in y = -1.25 m and the change in x = 20.0 m, but after trying a few things, I kept getting stuck. Please help! :smile:
 
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You have to set up two equations really. First you need to know how long it will take for the stunt rider to drop the 1.25 meters. Then you can work out how fast he needs to go to travel 20 meters horizontally in the time it takes for him to hit the ground again.
 
Oh, thank you so much. I hate it when the answer is staring me in the face. I appreciate the help!
 

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