Projectile Motion Involving Calculus

[TheAkuma]

Hi, I'm stuck in my maths assignment and need help with one of the questions.

“A stunt motorcyclist launches himself from a ramp inclined at 30 degrees to the horizontal. He aims to clear a line of cars that extends to a distance of 40 metres from the end of the ramp.
Use calculus methods to determine the minimum possible take-off speed for the rider given that the end of the ramp is 3 metres above the ground and an average car is about 1.5 metres tall.”


OK, so my teacher told me to treat the model as a coordinates on a Cartesian plane or something like that. At the bottom right of the ramp I put the origin there so i don't have to deal with the negative vertical axis. I used i for the horizontal axis and j for the vertical axis.
I have been given the equation v=v*√3/2 i+v/2 j from the ramp using sine and cosine, but I'm not sure if that's to work out the velocity at the point where the motorcycle leaves the ramp.

Now, I'm going to have a wild guess and say if I differentiate velocity I will get the displacement. But the problem with that is that it adds another variable, t, also i need to find c when i differentiate.

I need help finding the height (j) for the midpoint where the vertical velocity is zero. Since the question asks for the minimum velocity I was thinking that the midpoint would be at 20m or 20i. The path trajectory also needs to hit 40i+1.5j.

So my question is;
1.) How to find the constant (c) when differentiating velocity.
2.) How to find the height of the midpoint for the flight path.

If anyone could just point me in the right direction it would be very helpful. Also I have included two diagrams as attachments which I drew for the question.

 

[HallsofIvy]

Hi, I'm stuck in my maths assignment and need help with one of the questions.

So I have moved it to the homework section.

“A stunt motorcyclist launches himself from a ramp inclined at 30 degrees to the horizontal. He aims to clear a line of cars that extends to a distance of 40 metres from the end of the ramp.
Use calculus methods to determine the minimum possible take-off speed for the rider given that the end of the ramp is 3 metres above the ground and an average car is about 1.5 metres tall.”


OK, so my teacher told me to treat the model as a coordinates on a Cartesian plane or something like that. At the bottom right of the ramp I put the origin there so i don't have to deal with the negative vertical axis. I used i for the horizontal axis and j for the vertical axis.
I have been given the equation v=v*√3/2 i+v/2 j from the ramp using sine and cosine, but I'm not sure if that's to work out the velocity at the point where the motorcycle leaves the ramp.

Now, I'm going to have a wild guess and say if I differentiate velocity I will get the displacement.

Stop with the wild guesses! You should have learned that the derivative of velocity is acceleration. To find displacement you need to integrate.

But the problem with that is that it adds another variable, t, also i need to find c when i differentiate.

No, you will have a constant c, when you integrate. Apparently you are just mixing up the words "differentiate" and "integrate".

I need help finding the height (j) for the midpoint where the vertical velocity is zero. Since the question asks for the minimum velocity I was thinking that the midpoint would be at 20m or 20i. The path trajectory also needs to hit 40i+1.5j.

I'm not clear on what you are asking. Since the only force here is gravity (vertical), the horizontal velocity is constant. The minimum speed (not velocity) occurs when the vertical speed is 0.

So my question is;
1.) How to find the constant (c) when differentiating velocity.

When you integrate the velocity you get a "constant of integration" and I think that is the "c" you are asking about. Put a time and position that you know in the equation so you have only "c" left. Solve for c. You know the position of the motorcycle at the beginning, t= 0, don't you?[/quote]

2.) How to find the height of the midpoint for the flight path.

One method: having determined the beginning and ending points for the path, average their x-values to find the x-value of the midpoint. Put that into the equation for horizontal distance, x, as a value of t, and solve for t. Put that t into the formula for the vertical distance and calculate the height at that point.

Easier method: because of the symmetry, the midpoint will be the highest point in the path, when the vertical velocity is 0. Set the velocity function equal to 0 and solve for t. Use that t to calculate the height.

If anyone could just point me in the right direction it would be very helpful. Also I have included two diagrams as attachments which I drew for the question.

 

[TheAkuma]

thanks a lot, I've just been flooded with assignments I can't think straight.

This question doesn't really tell where the motorcyclist starts his run but it would be obvious that he starts a few meters behind the ramp. So I don't really know exactly where the displacement is at t=0, unless I state that he starts somewhere like 50m from the ramp to gain that minimum velocity. I have attached the equation I got from integrating velocity. Could you have a look to see if that's right? Also I don't think I'm allowed to use x and y coordinates as my teacher told me to use i's and j's (x=i), (y=j).