3) More trig:
Pythagorean Theorum
A^2 + B^2 = C^2
4) Still more trig:
Tan Θ = Opposite / Adjacent
Be sure your quadrant matches your initial vectors.
5) Probably need the figure. Sounds like an intercept problem. Watch your units on this one, they're mixing meters and miles.
6) Ditto.
7) Velocity * Time = Distance
8) Displacement = 1/2 a * t^2 + (Vo * t) Gravity is usually a negative.
Accel is ~ -9.8m/s^2 and Vo = 0 (ball was thrown horizontally, this is vertical displacement).
9) Vector problem.
Break the initial vector down into horizontal and vertical vectors and use the formulae from 7) and 8)
10)
If they want the same cliff from problem 7...
Since you know the initial vectors of the ball, and the time it takes to hit the ground...
you can do this:
y = 1/2 a * t^2 + (Vo * t)
Vo is the initial vertical vector. Plug and play, and it should spit out a negative distance as the result. The absolute value of that distance is the height of your cliff.