EASY Physics question?

Question

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the first one, you throw a second one at a low angle and timed to arrive at your opponent before or at the same time as the first one. Assume both snowballs are thrown with a speed of 30.0 m/s. The first one is thrown at an angle of 75.0° with respect to the horizontal.

(b) How many seconds later should the second snowball be thrown if it is to land at the same time as the first?
s

Answer 1

Hi,
Well, we know that for angles that exceed or fall short of 45 degrees (the angle for maximum range) by equal amounts, the ranges are equal. So, the angle of projection for the second snowball is:
θ = 45 – (75-45)
= 25 degrees.
So, let’s solve for “t” for the second snowball.
Now, we know that the displacement of the second snowball is:
S = (Vo cos θ)t (Eq#1) and if we knew S, we could solve for t.

Now, we can solve for the range of the first snowball (the second has to be the same or it won’t hit your “friend.”) from the this equation:
R = (Vo² sin 2θ)/g (Where g = 9.8 m/s²)
So, solve for that range and plug it into (Eq #1) in place of S and solve for t.

Now, for the time for the second snowball:
t = (Vo*sin θ)/g
So, plug the numbers into your calculator and solve for “ t” for that snowball.
Now, you have the time, t, for both snowballs, so subtract them and you’ll have the difference.
Just out of curiosity, if the problem is so EASY, why did you ask us for the solution?
:-)

Hope this helps.
FE.

Answer 2

I thought you said this one was easy.