a river current of constant velocity v. The helicopter is at a height of 9.8 m.

Question

a river current of constant velocity v. The helicopter is at a height of 9.8 m. The swimmer is 6.0 m na point directly under the helicopter when the life preserver is released. The life preserver lands 2.0 m 2a, 2b) In a rescue attempt, a hovering helicopter drops a life preserver to a swimmer being swept downstream by upstream fro in front of the swimmer . Neglect air resistance. Calculate the time it takes for the life preserver to hit the water after being released. a. b. Determine how fast the current is flowing.

Explanation / Answer

a)From the eqn of motion we know that,

S = ut + 1/2 at^2

In this case, S = H = 9.8 m; u = 0 ; a = g = 9.8 m/s^2 ; t = ?

9.8 = 0.5 x 9.81 x t^2

t = sqrt (9.8/0.5/9.8) = 1.414 s

Hence, t = 1.414 s

b)The swimmer travels,

d = 6 - 2 = 4 m in time t = 1.414 s

v = d/t = 4/1.414 = 2.83 m/s

Hence, v = 2.83 m/s

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