Posted by kia on Sunday, February 20, 2011 at 6:03am.
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, a second snowball is thrown at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 10.0 m/s. The first one is thrown at an angle of 65.0° with respect to the horizontal.
How many seconds later should the second snowball be thrown after the first to arrive at the same time?

physics  drwls, Sunday, February 20, 2011 at 9:51am
To arrive at the same PLACE, the second snowball must be launched at a complementary angle of 25 degrees.
The difference in the time of flight for those two launch angles is the required delay time, Tdelay.
Tdelay = 2 (Vsin65  Vsin25)/g
= [(10 m/s)/9.8 m/s^2]*(0.4836)
= 0.49 seconds 
physics  waqas, Saturday, August 6, 2011 at 7:44am
A particular Ferris wheel (a rigid wheel rotating in a vertical plane about a horizontal axis) at a local carnival has a radius of 20.0 m and it completes 1 revolution in 9.84 seconds.
(a) What is the speed (m/s) of a point on the edge of the wheel?
Using the coordinate system shown, nd:
the (b) x component of the acceleration of point A at the top of
the wheel.
the (c) y component of the acceleration of point A at the top of the wheel
the (d) x component of the acceleration of point B at the bottom of the wheel.
the (e) y component of the acceleration of point B at the bottom of the wheel
the (f) x component of the acceleration of point C on the edge of the wheel.
the (g) y component of the acceleration of point C on the edge of the wheel