Yep, it is a pretty routine motion Q, but, yeah, it's at the harder end, so I'll run you through it
(a) For average v, you will need displacement, s and time of flight, t. First up, split your calculations into two parts: finding a value for the maximum height reached; and, calculating the time of flight for both the upward, and the downward, leg.
v^2 = u^2 + 2as gives maximum height (for simplicity, treat all upward vectors as negative for calculations)
0 = (-6)^2 + (2 x 9.8)s
0 = 36 + 19.6s
s = -1.84m (2 D.P.)
Hence, maximum height reached is ~13.84m above the ground
t1 = (v - u)/a = [0 - (-6)]/9.8 = 0.61s (This is from v = u + at)
s = (at^2)/2 gives the time taken for the ring to drop from ~13.84m
Rearranging,
t2 = sqrt(2s/a) = sqrt[(2 x 13.84)/9.8] = 1.68s (as t scalar, only +sqrt is taken)
Now, t = t1 + t2 = 0.61 + 1.68 = 2.29s
Also, adding vectors: s = (-1.84) + (13.84) = 12m
So, v = s/t = 12/2.29 = 5.24m/s
i.e. the average velocity for the journey was 5.24m/s
(b) You already have time of flight and initial velocity. Since acceleration is given by a = (v-u)/t, finding the final (or impact) velocity will wrap this Q up for you.
final velocity, v, is found by considering the downward leg alone:
v = u + at = 0 + (9.8 x 1.68) = 16.46m/s
Now, a = [(16.46) - (-6)]/2.29 = 9.81m/s^2
(c) Nah, not going there, man. But, all this is asking for is that you graph acceleration, velocity and displacement against time.
Hope this helps out
And one more thing, don't girls usually go for the discreet dunk into the peaceful river from the bridge when they get screwed over? I could never imagine a girl throwing her engagement ring straight up from the top of a building like she was launching a rocket - but that's just me
