Projectile Motion

By Philip Ellison on Thursday, January 24, 2002 - 06:00 pm:

A projectile has velocity of projection 42m/s and passes through the point with position vector (200i+15j)m with respect to the point of projection. What are the two possible angles of projection to achieve this?
This should have a very simple solution... however, I can't see it.
Thanks

By Kerwin Hui on Thursday, January 24, 2002 - 08:58 pm:

I assume you take

j

to be the unit vector in the vertical upward direction and

i

to be a horizontal unit vector. Let the initial velocity be

42 (\cos θ i + \sin θ j)

This is a simple exercise of using 'suvat' equations: In the $i$ -direction, we have

$200 = 42 t \cos θ$

and in the $j$ -direction, we get

$15 = 42 t \sin θ - \frac{1}{2} g t^{2}$

Now just eliminate $t$ and solve the quadratic (remember that $\cos^{2} θ + \sin^{2} θ = 1$ ).

Kerwin

By Philip Ellison on Thursday, January 24, 2002 - 09:52 pm:

That's what I got... I just wasn't sure how to eliminate $t$

What are $(42 t \cos θ)^{2}$ and $(1 / 2 g t^{2})^{2}$ ?
Thanks

By Kerwin Hui on Thursday, January 24, 2002 - 09:59 pm:

We eliminate

t

by substituting

t = 100 / (21 \cos θ)

into

15 = 42 t \sin θ - \frac{1}{2} g t^{2}

. From

\cos^{2} θ + \sin^{2} θ = 1

, we deduce that

1 + \tan^{2} θ = \sec^{2} θ

, which is what you need to turn the equation into a quadratic in

\tan θ

Kerwin