p08_015.pdf

15. We use Eq. 8-18, representing the conservation of mechanical energy. We choose the reference position

for computing U to be at the ground below the cliﬀ;it is also regarded as the “ﬁnal” position in our

calculations.

(a) Using Eq. 8-9, the initial potential energy is U i = mgh where h =12 . 5mand m =1 . 50 kg. Thus,

we have

K i + U i = K f + U f

2 mv i + mgh =

2 mv 2 +0

which leads to the speed of the snowball at the instant before striking the ground:

v =

2 mv i + mgh = v i +2 gh

where v i =14 . 0 m/s is the magnitude of its initial velocity (not just one component of it). Thus

we ﬁnd v =21 . 0m/s.

(b) As noted above, v i is the magnitude of its initial velocity and not just one component of it;therefore,

there is no dependence on launch angle. The answer is again 21 . 0m/s.

the snowball does not change the result for v .