What a bad question. What do the drunken students have to do with any of this? And what's a daN?
Sigh...
Well it says horizontally so I guess that means over one's head. If so:
T sinθ = mg and I assume the daN thing is mg.
b)T cosθ = mv^2/r
You'll need to find r using a little trig, then solve for v
c) Now things get even murkier. Net? does he mean angular (change in omega over t)? Centripetal (v^2/r)? As for the angle do part b in reverse with the new (tangential I guess) velocity.
d) how did it suddenly get to 45o? Do we just ignore everything else and treat it as a straight projectile motion question? Who knows....
You see some management students getting drunk at 1 PM in the quad. (Yes, this is an early wake up
call for them.) You decide to tie a 2.00 daN water balloon on the end of a 1.00 m long light rope and
start swinging it horizontally. The rope from your hand to the water balloon makes an angle of 30.0
o
below the horizontal.
(a) What is the tension in the rope?
(b) What is the speed of the water balloon?
(c) Suppose you double the speed of the water balloon given in part (b) over four seconds. Assume
the acceleration is uniform. Just as you are approaching the four second mark, what is the net acceleration of the water balloon? What angle is the rope making with respect to the horizontal now?
(d) Suppose you release the rope and the water balloon flies off at the speed given in part (c) and at
an angle of 45.0 above the horizontal. What is the range of the water balloon if it left from a height of 1.75 m above the ground?
1 answer
1 answer