I would perfer if it was hand written please. 5.0 m/s Problem 3: Problem 3.52 in

Question

I would perfer if it was hand written please.

5.0 m/s Problem 3: Problem 3.52 in Young & Freedman An important piece of landing equipment must be thrown to a ship, which is moving at 45.0 cm/s, before the ship can dock. This equipment is thrown at 15.0 m/s at 60.0° above the horizontal from the top of a tower at the edge of the water, 8.75 m above the ship's deck (see the figure). For this equipment to 60.0P 45.0 cm/s 8.75 m the dock should the ship be when the equipment is thrown? Ignore air resistance

Explanation / Answer

let v = 45 cm/s = 0.45 m/s

vo = 15 m/s

theta = 60 degrees

voy = vo*sin(theta) = 15*sin(60) = 12.99 m/s
vox = vo*cos(theta) = 15*cos(60) = 7.5 m/s

let t is the time of flight

use, -h = voy*t + (1/2)*ay*t^2

-8.75 = 7.5*t + (1/2)*(-9.8)*t^2

t = 2.305 s

so,

D = (vox + v)*t

= (12.99 + 0.45)*2.305

= 30.8 m <<<<<<<<<---------------------Answer

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