1) A chitauritwo (Loki's weird alien friends) is flying past Hawkeye who is stationed on top of a building. The trajectory of the chitauritwo is given by the equations

x_c = -100m + 1ms^-1t + 1/2 0.1 t²

y_c = 100m

z_c = 440m

Hawkeye's position is given by x_c = 0, y_c = 0 and z_c = 440m. If he is to fire an arrow with an initial velocity of 137 ms^-1 at time t = 0, will he be able to shoot the chitauritwo? In what direction should he fire his arrow? Ignore air resistance, because the arrow is so fast, and note that gravity acts in the negative z-direction.

This is an intersection problem, so we have to write down the trajectory of the arrow and put the x-, y- and z-coordinates of the arrow as a function of time equal to the x-, y- and z-coordinates of the chitauritwo (given in the problem).

i. You have the starting location of the arrow so write down the equation for the position of the arrow in the x-direction as a function of time (note there is no air resistance so no acceleration). Put this equal to the x-coordinate of the chitauritwo to obtain your first equation.

ii. Do the same as in i but in the y-direction (should be simpler as the chitauritwo doesn't move in this direction).

iii. Obtain a similar equation in the z-direction. This is height so Hawkeye will have to aim slightly upwards to overcome the effects of gravity, which will appear in this equation as a negative term.

So now you have three equations, but 4 unknowns! The unknowns are the initial velocity of the arrow (in x, y, and z) and time. But wait, we have a fourth equation.

iv. The magnitude of the velocity is dictated by the bow. The magnitude is 137 m s^-1. Therefore, Pythagoras theory gives us our fourth equation. 4 unknowns and 4 equations. Now its a math problem. Plug it into a calculator or a computer if you can't do the math.

2) Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury. (Such a planet was once postulated, in part to explain the precession of Mercury's orbit. It was even given the name Vulcan, although we now have no evidence that it actually exists. Mercury's precession has been explained by general relativity.) The orbital period of Mercury is 88.0 days.

What would be the orbital period of such a planet?

You shouldn't need any help with this question as its a simple application of Kepler's 3rd law.

3) A skier leaves the ramp of a ski jump with a velocity of 10 m/s at 20^o above the horizontal as shown. The slope where she will land is inclined downward at 40^o, and air resistance is negligible.

Find the distance from the end of the ramp to where the jumper lands and her velocity components just before landing.

This is an interesting projectile motion problem, because the area where the projectile is landing (the projectile being the skier in this problem) is inclined.

i. First you need to write down the equation of motion for the skier in the x-direction after she has left the ski jump. Take the edge of the ski jump to be x = 0 and y = 0, and the point of time when she leaves as t = 0.

ii. Write down the equation of motion for the skier in the y-direction. Don't forget gravity.

iii. You should now have two equations but three unknowns (the x and y distances when she lands on the slope and the time this took). The third equation comes from the slope itself Given that the slope is going down at 40^o can you write an equation relating the x and y distances when she lands on the slope?