(I've posted this question on GameDev and I've been told to ask here so I'm pasting the question)

I need to make a car jump from a ramp to another, and I need this to be done using AI so I thought it would be enough to set a fixed speed for my car when near the ramp, in order to let it jump correctly.

It works fine if I input the speed manually but I'd like my game to calculate it since the ramps will be generated by user's input.

Unfortunately my physics knowledge lacks, so I need your help.

The problem is summarized in the picture below.

physics problem

Basically, what I need to compute is the speed (in m/s) that the car has to have on the highest point of the first ramp, given that all the data in red are user's input.

I had some rough ideas on using the projectile range with x=L, y=h-launchRampHeight but the results weren't as I expected them so I gave up.

Do you have any hints on this?

Please note that both ramps have the same length (50m) and that theta is expressed in degrees.


Based on the two replies I wrote some code and it seems to work fine but sometimes it returns a speed that's way higher than the correct one (which is not wrong but it's not the minimum one either).

For example, with theta = 8, L = 145, h = 12 it returns 100m/s (or maybe more, but the input's limited to 100m/s so it doesnt't show anything higher) while the jump works fine with 80m/s.

Can you please help me in sorting this out?


2 Answers 2


Well I tell you physically, you have to turn it to code yourself.

it's a simple projectile movement when we have:

$x =V_0tcos\theta $

$y=\frac{-1}{2} gt^2 + v_0tsin\theta $

where the $v_0$ is the initial velocity. and $\theta$ is the angle of thrown object.

if we want to get the equation without the time we will have:

$$y=\frac{-gx^2}{2V^2 \cos^2 (\theta)} + x\tan \theta$$

imagine that the place that the car is thrown (or jumped) the $(0,0)$ point. and $g$ is the acceleration that you defined in your engine (the gravity, I'm sure you know what the amount is in your engine). because you know the $\theta$ then all you need to do is to put the $(x,y)$ and calculate the velocity. before that I define a new variable $h'$ this way:

$h' =h- $**[height of my main ramp]**

and then put it in the main formula like this:

$$h'=\frac{-gL^2}{2V^2 \cos^2 (\theta)} + L\tan \theta$$

so the velocity will be equal to:

$$V=\sqrt{ \frac{-gL^2}{(h' - L\tan\theta) * 2 cos^2\theta}}$$

  • $\begingroup$ We have a sign difference, but I think it just comes from me playing with inequalities. $\endgroup$
    – zeldredge
    Apr 10, 2015 at 21:47
  • $\begingroup$ @zeldredge yeah, I didn't notice you answer. sorry. I guess we were answering the question at the same time. and yeah for the sign, since it is going to be used in that situation, I assumed the negative sign is not needed. and also I asked him in the comment, do you need the minimum velocity, and he said yes, So I didn't use an inequality. ;) $\endgroup$
    – Mobin
    Apr 10, 2015 at 21:56
  • $\begingroup$ Well, I actually had an $L$ and not an $L^2$ until you posted your answer, so thanks for helping me check units :p $\endgroup$
    – zeldredge
    Apr 10, 2015 at 21:57
  • $\begingroup$ @zeldredge ;) ;) $\endgroup$
    – Mobin
    Apr 10, 2015 at 21:58
  • 1
    $\begingroup$ @StepTNT Removing error is always the nasty part in these projects. the thing you have to do is to multiply a number to velocity like 0.8 to make it more real. the problem is that the lower velocity's will be less accurate then. you have to put if statement , and divide your results. for example if the velocity was between 100 and 80, then the velocity will be 0.8 . if you want it to be more smooth you have to make this multiplier a function. measure the error in each velocity and take the variance from it. then multiply the result on 1/variance. search the Google for more methods.take care ;) $\endgroup$
    – Mobin
    Apr 12, 2015 at 14:38

If you have initial velocity $v_0$ and initial launch angle $\theta$ (assumed $0 < \theta < \pi/2$), after a time $t$ you will have traveled through a distance: $$ \Delta x = v_x t = v_0 \cos \theta \\ \Delta y = - \frac{1}{2} g t^2 + v_y t = - \frac{1}{2} g t^2 + v_0 \sin \theta $$ The time required to traverse distance $L$ is the time you need to be in the air for: $$ t = \frac{L}{v_0 \cos \theta} $$ After this time we want our $\Delta y$ to be $h$ or better: $$ h < - \frac{1}{2} g \frac{L^2}{v_0^2 \cos^2 \theta} + \frac{L v_0 \sin \theta}{v_0\cos \theta}\\ <- \frac{1}{2} g \frac{L^2}{v_0^2 \cos^2 \theta} + L \tan \theta $$ Solve the inequality for $v_0^2$: $$ -\frac{1}{2} g \frac{L^2}{v_0^2 \cos^2 \theta} > h - L \tan \theta \\ v_0^2 > \frac{1}{2} g \frac{L^2}{( h - L \tan \theta) \cos^2 \theta}\\ v_0 > \sqrt{\frac{g L^2}{2 ( h - L \tan \theta) \cos^2 \theta}} $$ You, of course, want the positive root. Note that this also tells you when the problem is impossible: if $L \tan \theta > h$, then the straight-line path you get without falling at all isn't even enough to get you to the other ramp.

  • 1
    $\begingroup$ Comment: this is basically the first physics lab I ever did in high school, with a spring-loaded ball bearing launcher. It holds a very special place in my heart. $\endgroup$
    – zeldredge
    Apr 10, 2015 at 21:59
  • $\begingroup$ This is a though choice because you both replied with the same answer in the same time and I'd like to accept both of them but I can't. I edited the question with a little new issue tho :) $\endgroup$
    – StepTNT
    Apr 12, 2015 at 9:41

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