(a) How much later does it reach the bottom of the cliff?
using equation of motion (ii), we have
y = y0 + v0 t + (1/2) g t2
where, v0 = initial velocity of stone = 12.5 m/s
y0 = initial height of stone from top of the cliff = 0
y = final height of stone = - 75 m
g = acceleration due to gravity = - 9.8 m/s2
then, we get
- (75 m) = (0) + (12.5 m/s) - (0.5) (9.8 m/s2) t2
(4.9 m/s2) t2 - (12.5 m/s) t - (75 m) = 0
From quadratic equation, we get
t = - b (b2 - 4 a c) / 2a
t = - (-12.5 m/s) [(-12.5 m/s)2 - 4 (4.9 m/s2) (-75 m)] / [(2) (4.9 m/s2)]
t - 2.83947 sec OR t = 5.39049 sec
t = - 2.84 sec OR t = 5.40 sec
(b) What is its speed just before hitting?
using equation of motion (i), we have
v = v0 + g t
v = (12.5 m/s) - (9.8 m/s2) (5.40 s)
v = (12.5 m/s) - (52.92 m/s)
v = - 40.4 m/s
(c) What total distance did it travel?
using equation of motion (iii), we have
vf2 = v02 + 2 g d
(0)2 = (12.5 m/s)2 - 2 (9.8 m/s2) d
d = [(156.25 m2/s2) / (19.6 m/s2)]
d = 7.97 m
Then, we get
Dtotal = 2 d + y
Dtotal = [(15.94 m) + (75 m)]
Dtotal = 90.94 m
Dtotal 91 m
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