Question

Problem 2-
1. As a ship is approaching the dock at 45 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∘ above the horizontal from the top of a tower at the edge of the water, 8.75 m above the ship's deck as show in the figure. For this equipment to land at the front of the ship, at what distance D from the dock should the ship be when the equipment is thrown?

Answer 1

The equipment travels along a parabola such that its height at time t is given by

y = 8.75 + 13t - 4.9t^2
Solve for t when y=0 to see how long it will take to hit the deck.

Then, you know how far from the tower the package will hit, since its horizontal speed is 7.5 m/s.

Add that distance to the distance the ship travels in that time, to get D.

Answer 2

23.625

Answer 3

First identify the components of the velocity which are;
x = 15cos60, y = 15sin60

since there is height given we will find the time before it hits the water by using the formula:

-y=(vsin60∘)t-0.5(g)t^2

substitute the given values:
-8.75=(15sin60∘)t-0.5(9.8)t^2

as you can see we can use quadratic equation to find the value of t, or if you want to solve it in easiest way you can use your calculator:

solving the equation gives us t=3.209s
now we can solve for the D (range),
remember the distance formula which is X=vt, since we have the x component of the velocity, substitute V as Vcos60∘

x=(vcos60∘)t

substituting the following values gives us

D=24.075m

Answer 4

Since no one found the answer, I'll just put it here:

D= 25.5 m
is the final answer.

I still don't know how they found it. I tried every answer shown until I failed and they put the final answer without showing their work :(
Hope this helps even though I don't know how it was solved...