p04 072

background image

72.

(a) The helicopter’s speed is v

= 6.2 m/s. From the discussions in

§4-9 we see that the speed of the

package is v

0

= 12

− v

= 5.8 m/s, relative to the ground.

(b) Letting +x be in the direction of

v

0

for the package and +y be downward, we have (for the motion

of the package)

x = v

0

t

and

y =

1

2

gt

2

where ∆y = 9.5 m. From these, we find t = 1.39 s and ∆x = 8.08 m for the package, while ∆x

(for

the helicopter, which is moving in the opposite direction) is

−v

t =

8.63 m. Thus, the horizontal

separation between them is 8.08

(8.63) = 16.7 m.

(c) The components of

v at the moment of impact are (v

x

, v

y

) = (5.8, 13.6) in SI units. The vertical

component has been computed using Eq. 2-11. The angle (which is below horizontal) for this vector
is tan

1

(13.6/5.8) = 67

.


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