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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 qi + sin qj)

This is a simple exercise of using 'suvat' equations: In the i-direction, we have
200=42t cosq
and in the j-direction, we get
15=42t sin q-½gt2

Now just eliminate t and solve the quadratic (remember that cos2q+sin2q=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 (42tcosq)2 and (1/2gt2)2?
Thanks

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

We eliminate t by substituting t=100/(21 cos q) into 15=42t sin q - ½gt2. From cos2q+sin2q=1, we deduce that 1+tan2q=sec2q, which is what you need to turn the equation into a quadratic in tan q.

Kerwin