How Do You Calculate the Angle to Reach a Moving Target in a River?

[st0nersteve]

I was given this question as a study question but no solution provided. I am unsure how to solve the question.

Homework Statement

A man is swimming 3m/s down a river. The man is 50m from the edge and 1000m from the top of the river. If I am in a boat at the top corner of the river traveling 10m/s. what angle should i leave to reach the man.

The Attempt at a Solution

Currently i have no solution as to how to solve this equation. I am lost as to how to including movement of the swimming man into the equation

 

[ashishsinghal]

What do you feel? How can you start? Show some attempt first.

 

[st0nersteve]

Ive made no attempt because i have no clue what do do.
the formulas i have are the basic
v=u +at
v(2) = u(2) + 2as
s=ut + 1/2at(2)
Im lost because i don't know time, the angle or the distance because of the swimming guy moving.

Tried to find the velcoity of cos (downstream) by making the swimming guy 0m/s but then i got confused because that would make it 7cos(theta) but there's no side movemnt by the guy so it should still be 10sin(theta) across.

so i tried to sub them together in one equation but it didnit work out and I am not completely sure how to rearrange to solve for (theta).

my mess of working out was something like.
s(length) = ut + 1/2at(2)
1000=7cos(theta)xt + 0 50 = 10sin(theta) x t
t = 50/ 10 sin(theta)
therfore 1000 = 200 * 6cos(theta) / 8sin(theta)

Its completely screwed up i know. I am lost and my head hurts

 

[ashishsinghal]

Post a diagram and then explain your attempt.

 

[st0nersteve]

the scenario is the first diagram.

what i was thinking was diagram two. make the swimmer 0m/s. The only acting force for him is along the adjacent side, so the adjacent velocity will have to be 10cos$$\Theta$$ minus the 3m/s of the swimmer.

s = ut + 1/2at^2

Adjacent side Opposite side
50 = 10sin$$\Theta$$*t + 1/2*0*t^2 1000=(10cos$$\Theta$$ - 3)*t
50 = 10sin$$\Theta$$*t + 0 t= 1000/(10cos$$\Theta$$ - 3)
T=50/(10sin0$$\Theta$$)

Therefore: 50 = 10sin$$\Theta$$*(1000/(10cos$$\Theta$$ - 3))
50 = 1000*(10sin$$\Theta$$ / (10cos$$\Theta$$ - 3)

If this is anywhere near the right track I am lost here.
if that minus three wasnt there i know that 10sin0/10cos0 = 10tan0

hopefully you can help me because I am running out of time and don't know how to solve this