As a sailboat is approaching the dock at 45.0 cm/s, an important piece of landin

Question

As a sailboat is approaching the dock at 45.0 cm/s, an important piece of landing equipment needs to be thrown to it before it can dock. This equipment is thrown at 15.0 m/s at 60.0 o above the horizontal from the top of a tower at the edge of the water, 8.75 m above the boat’s deck.

1)How long does it take for the equipment to reach the level of the boat’s deck?

2)What is the distance the boat travels during that time?

3)At what distance from the tower would the equipment reach the deck’s level?

4)For this equipment to land at the front of the boat, at what distance D from the dock should the boat be when the equipment is thrown?

5)What are the equipment’s velocity components at the moment it hits the boat’s deck?

6)What is the maximum height the equipment will reach during its flight?

Explanation / Answer

1) in vertical,

yf - yi = v0y t + ay t^2 / 2

0 - 8.75 = (15 sin60) t + (-9.8 t^2 / 2)

4.9 t^2 - 13 t - 8.75 = 0

t =3.21 sec  

2) d = (45 cm/s)(3.21s) = 144.4 cm or 1.44 m  

3) dx = (v0x) t = (15 cos60) (3.21)

= 24.1 m

4) D = 24.1 - d = 22.6 m

5) vx = v0x = 15 cos60 = 7.5 m/s

vy = v0y +ay t = (15 sin60 ) + (-9.8 x 3.21)

vy = - 18.5 m/s

6) H_max = (15 sin60)^2 / (2 x 9.8)

= 8.61 m

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