Question

A sail boat approaches a dock at.500 m/s. A piece of landing equipment needs to be thrown into the ship before it arrives at the dock. The equipment is thrown at 12.5 m/s at an angle of 55 degree above the horizontal from the top of a light house that is 9 meters above the deck of the boat. What is the furthest distance (D) the boat sail can be and still catch the landing equipment? What is the maximum height that the landing equipment reaches during its flight? What is the magnitude of the velocity of the landing equipment when it reaches the ship? What is the angle with respect to the horizontal the landing equipment comes onto the ship with?

Answer 1

Using

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We get R = 19.76 m

time of flight =

= 2.756s

We can calculate maximum height reached by projectile above 9m scene using

h = v²sin²$\theta$/2g = 5.35 m

So maximum height above ground = 9+5.35 = 14.35m

Speed along vertical = v_{initial -}gt = 12.5sin 55 - 9.8*2.756 = -16.77m/s

Speed along horizontal = vcos$\theta$ = 12.5cos55 = 7.17 m/s

so v_net = 18.24m/s

For angle $\theta$ =tan^-1(16.77/7.17) = 66.85^o below horizontal