# Homework Help: Two motion issue

1. Feb 20, 2010

### barthayn

1. The problem statement, all variables and given/known data
A boat is 50.0m from the base of a cliff, fleeing at 5.0m/s. A gun, mounted on the edge of the cliff fires a shell at 40.0m/s and hits the boat when it has fled another 50.0m.

a) At what angel above the horizontal must the gun be aimed so that the shell will hit the target?
b) How high is the cliff?
c) With what velocity does the shell hit the boat?

Boat
v = 5m/s
d = 50m
t = 10s

Shell
vi = 40m/s
a = 9.8m/s2
t = 10s

2. Relevant equations
All the ones dealing with motion

3. The attempt at a solution
a)
vix = 100/10
vix = 10m/s

SinΘ = o/h
SinΘ = 10/40
SinΘ = 0.25
Θ = 14o

Therefore the angle about the horizontal must be 14o to hit the boat.

b)
vyf = vyi+aΔt
vyf = 38.7298+(10x9.8)
vyf = 38.7298+98
vyf = 136.73m/s

Δy = ((136.73+38.7298)/2)x10
Δy = 87.7x10
Δy = 877m
Δy = 880m

Therefore the cliff is 880m high.

c)

vf2 = 38.722+(2x9.8x880)
vf2 = 1500+17195
vf2 = 18695
vf = 136.73m/s
vf = 140m/s

x2 = 136.732+100
x2 = 19700
x = 140.357m/s
x = 140m/s

TanΘ = 140/10
TanΘ = 13.673
Θ = 85.818o
Θ = 86o

Therefore the shell hit the boat at 140m/s [86o below the horizontal]

Are these answers correct? I find the questions when two objects move at the same time very hard to do.

2. Feb 20, 2010

### aim1732

What is theta here? As far as I see vx=v*cos(theta)? [part a]

3. Feb 21, 2010

### barthayn

I got the angle from using sine. You do not need to use cosine because you have two sides and are looking for an angle. The angle I got was around 14o.

4. Feb 22, 2010

### aim1732

I don't get it. A diagram or something?

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