Stumped by problems dealing with "Curves and Motion in Space"?

Question

1) A projectile from the origin is to be fired at a target 600 feet away. There is a hill 300 feet high midway between the gun site and this target. At what initial angle of inclination should the projectile be fired?

*My guess to approach this is: Will this be a 2-part problem? Solving for an angle from origin to 300 ft then from 300 ft to 600? I don't know how to start exactly.

2) A point moves on a circle whose center is at the origin. Use the dot product to show that the position & velocity vectors of the moving point are always perpendicular.

*I know the answer is scalar, and must be 0 for the vectors to be perpendicular. That's as far as I can get for now. I will keep trying things till tomorrow morning (the last time I can check here before turning in these in).

Any helpful suggestions for either problem or both will be greatly appreciated. Thank you for your time and help! :)

Answer 1

The projectile's path will trace out a parabola with the vertex at coordinates (300,300)

Thus the general equation of the parabola is:

y-300 = a (x - 300)^2

The origin is one of the points on the parabola, thus

0-300 = a (0-300)^2
-300 = 900 a
a = -1/3

The specific equation is:

y - 300 = (-1/3) (x - 300)^2

Find the derivative of this equation at x=0 and set
tan Θ equal to the result.

2. Use polar coordinates

x = cos Θ
y = sin Θ

dx/dt = velocity in the x direction = - sin Θ * dΘ/dt
dy/dt = velocity in the y direction = cos Θ * dΘ/dt

The dot product is: (cos Θ) (- sin Θ) dΘ/dt + (sin Θ)(cos Θ) dΘ/dt = 0

Answer 2

If the target is on the same level as the origin, the range must be 300, that is

(V^2)(sin 2α)/g = 600 (if I remember the formulas correctly)

For a particular value of V, this will give two values of 2α between 0 and 180 deg, which means there are two values of α between 0 and 90 deg. We need to use the one which gives a maximum height greater than 300.

max ht = [(V^2)(sin α)^2]/(2g)

I can't just see how to get around not knowing V.

Answer 3

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