Projectile Motion question (1 Viewer)

[cutemouse]

A vertical wall, height h metres, stands on a horizontal ground. When a projectile is fired, in a vertical plane which is right angles to the wall, from a point on the ground c metres from the wall, it just clears the wall at the highest point of its path.

The equations of motion for the projectile with angle of projectile [maths]\theta[/maths] are:

[maths]x=Vtcos\theta \ y=Vtsin\theta-\frac{1}{2}gt^2[/maths]

(i) show that the particle reaches the highest point on its parth when [maths]t=\frac{Vsin\theta}{g}[/maths]

(ii) Show that the speed of projection is given by [maths]V^2 = \frac{g}{2h}(4h^2+c^2)[/maths]

(iii) Find the angle of projection, in terms of h and c.

Thanks

 

[daryl-d]

part 1, the derivate of the vertical displacement, ie y dot, sub y dot equals zero and the result follows

part 2, when max height occurs ie the value of 't' in part 1, x=c and y =h

sub these into horizontal and vertical displacements

rearranging vertical displacement u get

h = (v^2 sin^2 theta) divided by 2g

rerrage for sin^2 theta = 2gh/v^2

manipulating horizontal displacement

u get: C^2 = V^4 sin^2 theta Cos^2 theta / g^2 (obtain this result by squaring both sides of horizontal component)

then using trig identity ie. cos^2 theta = 1 - sin^2 theta

sub in the value for sin^2 theta from above into horizontal component

and just rearrage and the result follows

 

[Thecorey0]

Straight from the horse's mouth.

 

[random-1005]

A vertical wall, height h metres, stands on a horizontal ground. When a projectile is fired, in a vertical plane which is right angles to the wall, from a point on the ground c metres from the wall, it just clears the wall at the highest point of its path.

The equations of motion for the projectile with angle of projectile [maths]\theta[/maths] are:

[maths]x=Vtcos\theta \ y=Vtsin\theta-\frac{1}{2}gt^2[/maths]

(i) show that the particle reaches the highest point on its parth when [maths]t=\frac{Vsin\theta}{g}[/maths]

(ii) Show that the speed of projection is given by [maths]V^2 = \frac{g}{2h}(4h^2+c^2)[/maths]

(iii) Find the angle of projection, in terms of h and c.

Thanks

this is the 2009 catholic paper, its reasonably easy

 

[cutemouse]

I know how to do it... But I don't like the way the solutions set it out (part (ii) esp).

I was looking for a more 'direct' approach to this question.

 

[rheyn]

I took a direct approach to ii) by:
1. finding the expression for c and then the expression for h
2. substituting the c and h expressions into the RHS of

and then through simple algebra you can show that the
RHS = LHS

Also for part iii) you can also:
1. square both sides of c=Vtcos (theta)
2. then sub in v^2 and t^2 to get the answer in about 4 lines

Answer (using this method):