Posted by willy on Thursday, June 27, 2013 at 2:22pm.
Batman is standing on a cliff 120 m high. He is testing a new bat grappling hook. He
fires the hook with an initial velocity of 85 m/s at an angle of 45o
with respect to the
horizontal. How far from the base of the cliff (d) does the hook land?

physics  Henry, Friday, June 28, 2013 at 8:43pm
Vo = 85m/s[45o].
Xo = 85*cos45 = 60.1 m/s.
Yo = 85*sin45 = 60.1 m/s.
Tr = (YYo)/g = (o60.1)/9.8 = 6.1 s.
= Rise time.
h = Yo*t + 0.5g*t^2
h = 60.1*6.1  4.9*(6.1)^2 = 184.3 m.
d = Yo*t + 0.5g*t^2=184.3 + 120=304.3m
0 + 4.9t^2 = 304.3
t^2 = 62.1
Tf = 7.9 s. = Fall time.
d =Xo * (Tr + Tf)
d = 60.1*(6.1+7.9) = 841.4 m.
91,75s = 105 m.
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