Gene C. answered 11/01/15
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I see that no one has answered your question yet. This is a common problem in physics and dynamics problems in engineering.
a) The basic formula is y=v0t-gt2/2
9.81m/sec2 x t2 - (14 m/sec x t) - 60m = 0
quadratic equation t= -b+/- √b2-4ac /2a
t= 14+- √(-14)2 -4(4.905)(-60)/ 2(4.905) = 5.20 sec
b) y = v0t - gt2/2
At max height H, Vy = 0
Vy2 = V20y - 2g(y-y0)
0=(14)2 -2(9.81)(H-60)
H=70m
The ball would fall 10m from max height of 70m. set y=0 and H=10m
y=gt2/2+H, t = 1.42 sec.
Therefore, 5.2 - 1.42 = 3.77 sec

Hilton T.
It takes (14 - 0)/ 9.81 = 1.42 s to reach maximum height. It will therefore take a further 1.42 s to return to the point from which it was launched.
Total time = 2 x 1.42 = 2.84 s
11/03/15