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.
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.
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 = ????