1. urgent help needed!

ok so ive been up all night at this but I dont no what to do!
so here goes:
A projectile released with a velocity U m/s at an angle xto the horizontal, just clears two obstacles, both of height h m, whose distances from the point of projection are b m and 3b m respectively. Find the range of the projectile.
any help or answers would be greatly appreciated. Thanks.

2. Originally Posted by totster
ok so ive been up all night at this but I dont no what to do!
so here goes:
A projectile released with a velocity U m/s at an angle xto the horizontal, just clears two obstacles, both of height h m, whose distances from the point of projection are b m and 3b m respectively. Find the range of the projectile.
any help or answers would be greatly appreciated. Thanks.
Let the angle to the horizontal be $\displaystyle \theta$.

$\displaystyle y=xtan\theta -\frac{gx^2}{2u^2cos^2\theta}$

Since the projectile passes through the points (h, b) and (h, 3b), we get

$\displaystyle b=htan\theta -\frac{gh^2}{2u^2cos^2\theta}$

$\displaystyle 3b=htan\theta-\frac{gh^2}{2u^2cos^2\theta}$

Solve for $\displaystyle u$ and $\displaystyle \theta$ from these equations and substitute in

$\displaystyle R=\frac{u^2sin2\theta}{g}$

3. Originally Posted by totster
ok so ive been up all night at this but I dont no what to do!
so here goes:
A projectile released with a velocity U m/s at an angle xto the horizontal, just clears two obstacles, both of height h m, whose distances from the point of projection are b m and 3b m respectively. Find the range of the projectile.
any help or answers would be greatly appreciated. Thanks.
Integrate the equations of motion to get the horizontal and vertical displacements as functions of time.

From the horizontal displacement equation find t in terms of horizontal displacement and substitute this into the vertical equation.

This gives you a quadratic for the vertical displacement in terms of the horizontal displacement.

Now use this to complete the solution.

CB

4. The flight path will be symetrical about a vertical line at some distance x. But we know that the projectile is at the same height when it is at b and at 3b. Therefore the line of symetry must be located half way between b and 3b, that is at 2b.

Now you know the projectile is at height 0 m at x=0 and it has a line of symetry at x=2b. So the projectile will again be at height =0 when x = ????