Optimizing Boat Angle for Hitting a Target Across a River

AI Thread Summary
To hit a target 75m across a 400m wide river with a boat affected by a constant eastward current of 0.5 m/s, the angle north of west needs to be calculated. The initial equations set up for the problem contain errors in the treatment of velocities and trigonometric functions. The correct approach involves using the relationship between the boat's velocity components and the river's current, leading to a goniometric equation that can be solved for the angle. It's essential to clarify units and ensure proper mathematical relationships are maintained to avoid nonsensical results. Properly addressing these issues will enable the calculation of the required angle to successfully hit the target.
KurtWagner
Messages
43
Reaction score
1
So I have a boat going across a river (y-direction, north) 400m wide.
I am trying to hit a target 75m in the positive x direction on the opposite side of the river.
My boat will travel a velocity v.
The river will provide a constant velocity of +0.5 (east)

I am trying to find the angle a, north of west to point my boat.

hope that all makes sense.

so far I have:

y = vt or t = y/v
t = 400 / v*sin(a)

x = vt
75 = [0.5 - cos(a)] * t

then I plugged in t

75 = 200/[v*sin(a)] - [400*v*cos(a)]/[v*sin(a)]

I can't seem to solve this. Am I doing this the right way?

I was thinking I need another function and I remember having to use
sin(a)^2 + cos(b)^2 = 1
for something like this a while back. And if I do, how do I? Do I just solve the first two equations for sin and cos and then plug the equations into the third, expand them, then solve another for t and then plug it in and simplify?

any help would be greatly appreciated.
 
Physics news on Phys.org
Hello Kurt, sorry you missed the template. Please use it.

I wonder if your river velocity is 0.5 nautical miles per fortnight or 0.5 m/s, because you missed giving the units. With an error as a consequence:

If you had written your second relevant formula (under 2. in the template) as

75 m = [0.5 m/s - cos(a)] * t

you would have immediately seen that it is nonsense to add a cosine to a velocity (how long is a year plus one ?)

However, this appears to be a typo omission, because under "3. attempt at solution", the v pops up again.

A second error pops up if you temporarily assume the river not to flow. a can then be zero, but instead of t = 400 m / v you get a divergence. You want to interchange cos and sin.

For the rest, you're okay:
vt cos(a) = 400 m plugged into the second eqn gives you
$${75\over 400}v\cos\alpha = 0.5 - v \sin\alpha$$
which is a simple goniometric equation of the form ##a \cos\alpha + b\sin\alpha = c##.
The relevant expression to help you solve that is ##\sin(\alpha+\beta) = ...##
 
thanks. sorry for missing the units and for the typo also. I'll give it another crack. :)
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

Similar threads

Back
Top