Cannon shooting emergency packets

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Homework Help Overview

The problem involves a cannon shooting emergency packets to a roof of a building that is 120 meters high and located 40 meters away from the cannon. The goal is to determine the initial speed and angle of the cannon so that the packets land gently on the roof.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationships between the equations of motion and the variables involved, such as initial velocities and angles. There are attempts to express one variable in terms of others and to substitute values into the equations. Questions arise regarding the setup of the problem and the interpretation of the conditions for landing.

Discussion Status

Participants are actively engaging with the problem, sharing their attempts and reasoning. Some guidance has been offered regarding the approach to take with the equations, and there is acknowledgment of progress made by some participants in understanding the relationships between the variables.

Contextual Notes

There is mention of a diagram that may clarify the problem setup, and participants are navigating through the complexity of having three equations with three unknowns, which adds to the challenge of finding a solution.

naianator
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Homework Statement


A cannon at ground level is shooting emergency packets to people stranded on the roof of a flooded building of height H=120 meters. The corner of the building is located a distance D=40 meters from the cannon. It is desired that the incoming packets are flying tangent to the roof as shown so that they land gently with as little impact as possible and slide along to a stop.

Find the initial speed v0 and at what angle θ (in degrees) the cannon should be aimed to achieve the above scenario.

Homework Equations


x(t)=x_0+v_0x*t+1/2*a*t^2

The Attempt at a Solution


x(t)=v_ox*t
40=v_ox*t
t=40/v_ox
y(t)=v_oy*t-1/2*g*t^2
120=v_oy*t-1/2*g*t^2
v_y(t)=v_oy-g*t and since it hits the roof at the top of the curve v_y(t)=0=v_oy-g*t and
t=v_oy/g
This is where I get stuck. v_oy/g=40/v_ox and i can't find a way to relate the velocities so that i can substitute tan(theta)=v_oy/v_ox

Same goes if i try to substitute t=40/v_ox into 120=v_oy*t-1/2*g*t^2 so maybe I'm just going about this the wrong way

Please help
 
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naianator said:
... tangent to the roof as shown ...
Shown where?
 
You have three equations and three unknowns. The basic technique is to find or develop an equation which expresses one unknown in terms of the other two, then use this to replace that unknown in each of the other two equations. You will then have two equations with two unknowns. Repeat process.
 
haruspex said:
You have three equations and three unknowns. The basic technique is to find or develop an equation which expresses one unknown in terms of the other two, then use this to replace that unknown in each of the other two equations. You will then have two equations with two unknowns. Repeat process.
Thats sort of what I've been trying to do. Any suggestions as to which one to use?
 
i found v_ox=40/t so v_ox=40g/v_oy and v_oy=v_ox*tan(theta) so v_ox=sqrt(40g/tan(theta)) but I'm not sure how to relate that back to v or if it even helps me

and i think the picture should be attached now
 

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naianator said:
Thats sort of what I've been trying to do. Any suggestions as to which one to use?
You had v_oy/g=40/v_ox, and you mentioned substituting t=40/v_ox into 120=v_oy*t-1/2*g*t^2, but didn't post the result.
Having made those two steps, you should be down to two equations and two unknowns.
 
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haruspex said:
You had v_oy/g=40/v_ox, and you mentioned substituting t=40/v_ox into 120=v_oy*t-1/2*g*t^2, but didn't post the result.
Having made those two steps, you should be down to two equations and two unknowns.
Yes, I got it thank you. I was overlooking a couple of things.
 

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