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 = (Y-Yo)/g = (o-60.1)/-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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