p04 086

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86.

(a) From Eq. 4-22 (with θ

0

= 0), the time of flight is

t =



2h

g

=



2(45)

9.8

= 3.03 s .

(b) The horizontal distance traveled is given by Eq. 4-21:

x = v

0

t = (250)(3.03) = 758 m .

(c) And from Eq. 4-23, we find

|v

y

| = gt = (9.80)(3.03) = 29.7 m/s .


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