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A helicopter drop say supply package to to flood victims on a raft in a swollen lake. When the package is released it is 88 m directly above the raft and flying due east at 78.3 mph, a) how long is the package in the air, b) how far from the raft did the oackege land c)what is the final velocity of the package

User Mish
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We can use the equations of motion to solve this problem.

a) 4.1 seconds

- We need to find the time it takes for the package to land on the raft. The initial vertical velocity is zero, and the acceleration due to gravity is -9.81 m/s^2 (negative because it opposes the upward motion).

We can use the equation:

h = vt + (1/2)at^2

where h is the initial height (88 m), v is the initial vertical velocity (zero), a is the acceleration due to gravity (-9.81 m/s^2), and t is the time.

Plugging in the values, we get:

88 = 0 x t + (1/2)(-9.81)(t^2)

Simplifying and solving for t, we get:

t = sqrt((2 x 88)/9.81)

t ≈ 4.1 seconds

Therefore, the package is in the air for 4.1 seconds.

b) 1.25 km

- We need to find the horizontal distance travelled by the package in 4.1 seconds. The initial horizontal velocity is 78.3 mph (we convert to m/s), and the acceleration is zero (since there is no horizontal force acting on the package).

We can use the equation:

d = vt

where d is the distance, v is the initial horizontal velocity, and t is the time.

Plugging in the values, we get:

d = 78.3 mph x (1.609 km/m)(1/3600 h/s) x 4.1 s

d ≈ 1.25 km

Therefore, the package lands about 1.25 km east of the raft.

c) 97.5 m/s

- We can use the components of velocity to find the final velocity of the package. The vertical velocity is -gt, where g is the acceleration due to gravity and t is the time of flight (4.1 seconds). The horizontal velocity is 78.3 mph (which we convert to m/s).

The final velocity can be found using the Pythagorean theorem:

vf = sqrt(vh^2 + vv^2)

where vh is the horizontal velocity and vv is the vertical velocity.

Plugging in the values, we get:

vf = sqrt((78.3 mph x (1.609 km/m)(1/3600 h/s))^2 + (-9.81 m/s^2 x 4.1 s)^2)

vf ≈ 97.5 m/s

Therefore, the final velocity of the package is about 97.5 m/s at an angle of tan^-1(-(9.81 m/s^2 x 4.1 s) / (78.3 mph x (1.609 km/m)(1/3600 h/s))) = -0.134 rad = -7.7 degrees below the horizontal.

User Tim Rutter
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