Physics question?

Question

A draw bridge over a river has two horizontal halves that raise when needed. In the raised position, each section is 40 degrees to the horizon. A car driver gets to the bridge in the open position. Wanting to clear the gap, the driver guns the car and leaves the ramp at 36 m/s. I figured out that the driver does clear the gap, but I also have to figure out where the driver lands on the other section of the ramp (cant seem to get it).

Thanks in advance.

Answer 1

Let's set the total width of the river as w, so the length of the bridge half is w/2.

The horizontal speed of the car is
cos(40)*36

The vertical is
sin(40)*36

The expression of the vertical location is
y=sin(40)*36*t-.5*g*t^2

You also know that
x=cos(40)*36*t

Since you didn't provide the width of the river, I can't take the solution all the way, so I'll describe the technique.

Set y=0 and compute the two roots of the equation.

One will be zero, since that was the starting point
the other will give you the time that the car falls back below the height of the river side of the raised ramp that is the landing side.

Check the x location and that was how you proved that the car had traveled a distance greater than w*(1-cos(40))

Once past the end, you now look for where the car descent intersects the ramp.

For (w*(1-cos(40)/2)>
x>
w*(1-cos(40))

the surface of the landing ramp below the height of the end of the launching ramps is y for the landing location
y=
-(w/2*sin(40)
-tan(40)*
(w/2-x+w/2*(1-cos(40))
set this equal to y for the car
sin(40)*36*t-.5*g*t^2

replace x with
x=cos(40)*36*t

solve for the greater t and you have it

you should test to make sure the car hits the ramp.

Look at t for x=(w*(1-cos(40))
and then check the height of the car. If it calculates more than
0, then it clears the top of ramp and lands on the ramp.


This is an update:

I did all of the algebra and set up the equations in Excel to look at various river widths.

The solution requires solving a quadratic equation to find where the trajectory of the car intersects the ramp on the landing side.
I will just show the coefficients:

a*t^2+b*t+c=0
a=-9.81/2
b=36*(sin(40)+tan(40)*cos(40))
c=w*(sin(40)-
tan(40)*(2-cos(40)))/2

Keep in mind that the problem is piecewise linear, so one root will be outside the range of x where the y value is valid.

It turns out the root that matters is
(-b-sqrt(b^2-4*a*c))/(2*a)

for a river that is 350 meters wide, the car lands on the ramp
at a distance of 6.6 meters before the end of the bridge at
t=7.6 seconds after leaving the first ramp.

If x>(w*(1-cos(40)/2)

Then it lands on the road past the bridge

For this case, you calculate the intersection of
y=-sin(40)*w/2
=sin(40)*36*t-.5*g*t^2

This is much easier to calculate. Again, it is a quadratic.

For a bridge 100 m long the car flies for 5.84 seconds and lands 99 m past the end of the bridge.

j