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(cosqi + sinqj)

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 sinq- 1
2
 g t2

Now just eliminate t and solve the quadratic (remember that cos2 q+sin2 q = 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 cosq)2 and (1/2g t2)2?
Thanks

By Kerwin Hui on Thursday, January 24, 2002 - 09:59 pm:
We eliminate t by substituting t=100/(21cosq) into
15=42tsinq- 1
2
 g t2

. From cos2q+sin2q = 1, we deduce that 1+tan2q = sec2q, which is what you need to turn the equation into a quadratic in tanq.

Kerwin