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

AI Thread Summary
The discussion revolves around calculating the minimum speed required for a stunt driver to jump a car 21.1 meters from a ramp 1.48 meters high. Initial calculations involved applying kinematic equations, with a focus on determining time and initial velocity. After identifying an error, the correct time for the jump was found to be approximately 0.5496 seconds, leading to a calculated initial velocity of about 35.7 m/s. The second part of the problem involves adjusting the calculations for a ramp tilted at an angle of 11.6 degrees, prompting a comparison to projectile motion. The conversation emphasizes the importance of understanding kinematics and projectile dynamics for solving such problems.
hotmail590
Messages
52
Reaction score
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:
Physics news on Phys.org
Yes,u can start by applying the laws of kinematics of movement with (presumably) constant acceleration in free fall.

Post some ides,equations...

Daniel.
 
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:
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.
 
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
 
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top