Solve Two Motion Issues: Boat & Shell Hit Target

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Homework Help Overview

The problem involves two objects in motion: a boat moving away from a cliff and a shell fired from a gun on the cliff. The objective is to determine the angle at which the gun must be aimed, the height of the cliff, and the velocity of the shell upon impact with the boat.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to calculate the angle using trigonometric functions and provides values for the boat's and shell's velocities and distances. Some participants question the use of trigonometric functions, particularly regarding the angle calculations.

Discussion Status

Participants are exploring different interpretations of the angle calculations and the relationships between the velocities and angles involved. There is a request for clarification through diagrams, indicating a need for visual representation to aid understanding.

Contextual Notes

The original poster expresses difficulty with problems involving simultaneous motion, highlighting a potential area of confusion in the discussion.

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Homework Statement


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


Homework Equations


All the ones dealing with motion


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.
 
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What is theta here? As far as I see vx=v*cos(theta)? [part a]
 
aim1732 said:
What is theta here? As far as I see vx=v*cos(theta)? [part a]

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