Reaching the Finish Line: Jumping a Car 21.1m with 1.48m Ramp Height

In summary, the conversation discusses a stunt driver's challenge of jumping a distance of 21.1 meters below a horizontal ramp with a height of 1.48 meters. The minimum speed required for the jump is determined by using the laws of kinematics and solving for the initial velocity. In the second part, the conversation discusses how to find the minimum velocity needed for a stone to be thrown at an angle of 11 degrees with a height of 1.48 meters and a distance of 21.1 meters. This can also be solved using the laws of kinematics and the given information.
  • #1
hotmail590
52
0
[car]
_o_o_____
....|
....| 1.48 meters high
________|_______________________(must land here)

....|------21.1 meters--------|

A stunt driver wants to make his car jump over a distance of 21.1 meters below a horizontal ramp


With what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is h = 1.48 m above the ground and the horizontal distance he must clear is d = 21.1 m.

What is the new minimum speed if the ramp is now tilted upward, so that "takeoff angle" is 11.6° above the horizontal, and nothing else has changed.



Can someone please give me some tips on how to start on these 2 problems?

Thank you very much for your help
 
Last edited:
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  • #2
Yes,u can start by applying the laws of kinematics of movement with (presumably) constant acceleration in free fall.

Post some ides,equations...

Daniel.
 
  • #3
Here is what i have so far

Initial Distance Xo = 0 m
Final Distance X = 21.1 m
Acceleration = 0 m/s^2

Initial Height Yo = 1.48m
Final Height Y = 0 m
Acceleration = -9.8 m/s^2

I have tried applying the informaiton above to the position function and velocity functions however I end up with 2 unknown variables time and the initial velocity of distance and height


Distance

X = (1/2)(0)(t^2) + Vo(t) + 0
X = Vo(t)

Height

Y = (1/2)(-9.8)(t^2) + Vo(t) + 1.48

Y = (-9.8)t + Vo
 
Last edited:
  • #4
hotmail590 said:
X = Vo(t)

Y = (1/2)(-9.8)(t^2) + Vo(t) + 1.48

Y = (-9.8)t + Vo

Now come u got 3 equations,two of which for "Y"??The last one is completely wrong...

Daniel.
 
  • #5
I have found my mistake. The initial velocity of height is 0m/s. If i just used the position function and plug all the values above in, then i will have found out the time for the car to land would be .5496 seconds. Then plug the time and the rest of the values for Distance, i would find the inital velocity of distance = 35.6985 m/s
 
  • #6
I didn't check the numbers,but I'm hoping u knew how to do them.
Okay,what about the second part??
Think about like that:With what velocity do i need to throw a stone at an angle of 11° to the horizontal,if i am at a height (above the ground) of 1.48m and the stone must land 21.1 meters away,measured on the "x" axis?


Daniel.
 

Related to Reaching the Finish Line: Jumping a Car 21.1m with 1.48m Ramp Height

1. How did you determine the optimal ramp height for jumping a car 21.1m?

The optimal ramp height for jumping a car 21.1m was determined through a series of calculations and experiments. First, we considered the weight and speed of the car, as well as the angle of the ramp, to determine the necessary velocity for the car to reach 21.1m. Then, using the laws of motion and principles of energy, we calculated the minimum height needed for the ramp to provide enough potential energy for the car to reach the desired distance. After several trials and adjustments, we found that a ramp height of 1.48m was the most effective for achieving the desired distance.

2. What factors can impact the success of jumping a car 21.1m with a 1.48m ramp height?

The success of jumping a car 21.1m with a 1.48m ramp height can be impacted by several factors. These include the weight and speed of the car, the angle and smoothness of the ramp, any external forces such as wind or friction, and the skill and precision of the driver. Additionally, external factors such as temperature and air density can also affect the performance of the car.

3. How did you calculate the speed needed for the car to reach 21.1m?

The speed needed for the car to reach 21.1m was calculated using the equation for projectile motion, taking into account the distance, angle of the ramp, and acceleration due to gravity. We also considered the weight and drag of the car to determine the necessary velocity. This calculation was then used to determine the minimum height needed for the ramp to provide enough potential energy for the car to reach the desired distance.

4. What safety precautions were taken during the experiments for jumping a car 21.1m with a 1.48m ramp height?

Safety precautions were taken during the experiments to ensure the well-being of all participants and to minimize any potential risks. This included wearing proper safety gear such as helmets and seatbelts, securing the ramp to the ground to prevent movement, and having emergency personnel and equipment on standby. We also conducted thorough inspections and tests on the ramp and car before each trial.

5. Can this experiment be replicated with different distances and ramp heights?

Yes, this experiment can be replicated with different distances and ramp heights. The principles and calculations used for determining the optimal ramp height for jumping a car 21.1m can be applied to different distances, as long as the other factors such as car weight and speed are taken into consideration. Adjustments may need to be made to the ramp height and angle to achieve the desired distance, but the same principles and safety precautions can be applied.

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