Physics ASAP!!!!!!!!!!!! Please Help!!!!!!!!!?

Question

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. Then, while your opponent is watching that snowball, you throw a second one at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown at 25.0 m/s. The first is thrown at an angle of 70 degrees with respect to the horizontal.
a) At what angle should the 2nd snowball be thrown to arrive at the same point as the 1st?
b) How many seconds later should the second snowball be thrown after the 1st in order for both to arrive at the same time?

the answers:
a) 20 degrees
b) 3.05 sec

Can you tell me how to do it? PLEASE!?

Answer 1

An object thrown straight upwards at velocity v0 will hit the ground with velocity -v0, which requires a total velocity change of 2v0. The amount of time required for this to take place is 2v0/g.

If you throw an object at angle theta to the horizontal with initial velocity v0 then its vertical velocity component vv will be equal to v0 * sin(theta) and its horiziontal velocity component vh will be v0 * cos(theta).

The object will hit the ground after time 2vv/g, at which point it will have travelled (2vv/g)*vh horizontally. This works out to 2 * v0^2 * sin(theta)*cos(theta)/g, which simplifies to v0^2/g * sin(2*theta). This is consistent with the fact that a 45 degree throw will travel the farthest distance.

So, if both snowballs are thrown at the same initial velocity, it follows that the sin(2*theta) is the only factor that changes. In the case of the first snowball we have sin(140). This turns out to be the same as sin(40), so a snowball thrown at 20 degrees will travel the same distance.

The first snowball will take 2vv/g time to hit the ground, or 2 * 25 * sin(70) / g. The second snowball will take 2 * 25 * sin(20) / g. Work these out, and find the difference, your answer to question (b).