Question

It is required to project a body from a point on level ground in such a way

as to clear a thin vertical barrier of height h placed at a distance a from the

point of projection. Show that the body will just skim the top of the barrier

if

ga2

2u

2



tan2 α − a tan α +

ga2

2u

2

+ h



= 0

where u is the speed of projection and α is the angle of projection above the

horizontal

Answer 1

Suppose that the motion starts from the origin and takes place in the .x; z/-plane,

where Oz points vertically upwards. Then the path of the body

$$z=x\tan \alpha-\frac{g}{2u^2\cos^2{\alpha}}x^2$$

where u is the projection speed and $\alpha$ is the angle between the direction of projection and the positive x-axis. If the path just skims the top of the

barrier, then u and $\alpha$ must satisfy the equation

$$h=a\tan \alpha-\frac{g}{2u^2\cos^2{\alpha}}a^2$$

On using the trigonometric identity

$$\sec^2\alpha=1+\tan^2\alpha$$

this condition can be written in the form

$$ga^2\tan^2\alpha-2au^2\tan \alpha+(ga^2+2hu^2)=0$$