Projectile motion (1 Viewer)

Kingportable · Jul 22, 2012

A vertical wall, height h meters, stands on horizontal ground. When a projectile is fired, a vertical plane which is at right angles to the wall, from a point on the ground c meters from the wall, it just clears the wall at the heightest point on its past. The equations of motions of motion for the projectile with angle of projection, theta, are:

X=vosTHETAt" Y=vsinTHETAt - 1/2.gt^2

Hegihest point of projection is t=vsinTHETA/g

1. Show that the speed f projection is given by v^2= g(4h^2+c^2)/2h
2. Find the angle of projection , THETA, in terms of h and c.

This one has confused with question one I initially went to subbing to y and saying that At dy/dt=0 y=h and by us being t into to. The horizontal motion x would equal h.

Sy123 · Jul 22, 2012

$x=vt\cos{\theta} \\ y=vt\sin{\theta}-\frac{1}{2}gt^2 \\ \\ $when$ \ \ t=\frac{v\sin{\theta}}{g} \ \ \ y=h \ \rightarrow x=c \\ \\ \therefore c=\frac{v^2 \sin{\theta} \cos{\theta}}{g} \\ h=\frac{v^2 \sin^2{\theta}}{2g}$

I substituted the time when x=c and y=h (the highest point of the projection). I am now going to manipulate both equations of both c and h, into a form, where I can construct the expression in the question given.

$c^2=\frac{v^4 \sin^2{\theta} \cos^2{\theta}}{g^2} \ ... \ \ [1] \\ \\ 2h=\frac{v^2 \sin^2{\theta}}{g} \ ... \ \ [2] \\ \\ h^2=\frac{v^4 \sin^4{\theta}}{4g^2} \ ... \ \ [3]$

Now to form an expression using equations 1, 2 and 3. We will see what we can get.

$\frac{g}{2h} (4h^2+c^2) \ \ $substitute [1] and [2] and [3]$ \\ \\ =\frac{g}{\frac{v^2 \sin^2{\theta}}{g}} \left (4(\frac{v^4 \sin^4{\theta}}{4g^2})+\frac{v^4 \sin^2{\theta} \cos^2{\theta}}{g^2} \right) \\ \\ =\frac{g^2}{v^2 \sin^2{\theta}} \left (\frac{v^4 \sin^2{\theta}(sin^2{\theta}+\cos^2{\theta}}{g^2} \right ) \\ \\ =v^2$

I skipped some of the algebra, simply combine the fractions then take out common factors

PART II

$h=\frac{v^2 \sin^2 {\theta}}{2g} \ \ ... \ \ ($From part i$) \\ \\ 2gh=\frac{g}{2h}(4h^2+c^2) \sin^2{\theta} \\ \\ sin^2{\theta}=\frac{4h^2}{4h^2+c^2} \\ \\ \therefore \theta=\sin^{-1}{\sqrt{\frac{4h^2}{4h^2+c^2}}}$

I substituted the v^2 form from the first part into one of the equations. Then I rearrange to find theta.

Kingportable · Jul 22, 2012

You sir are a champion.

Projectile motion (1 Viewer)

Kingportable

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Sy123

This too shall pass

Kingportable

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