Can Physics Save the Hospital from Mortar Fire?

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Homework Help Overview

The problem involves two groups firing mortars at a children's hospital, with one group attempting to intercept the other's projectile. The scenario includes projectile motion calculations, with specific angles and distances provided. The context is rooted in kinematics and the physics of motion under gravity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the time it takes for projectiles to reach the hospital and the necessary initial velocities for interception. There are attempts to equate the vertical positions of both projectiles and questions about the role of time in their calculations.

Discussion Status

Some participants have offered guidance on setting up equations to relate the velocities and time, while others express uncertainty about the correct approach. Multiple interpretations of the problem's requirements are being explored, particularly regarding the timing and positioning of the projectiles.

Contextual Notes

There is a noted confusion about whether the projectiles need to hit the hospital simultaneously or if one must intercept the other mid-flight. Participants are also grappling with the implications of time in their equations and the relationships between the variables involved.

lynnx95
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A group of elderly, hospital-hating street toughs fire a mortar at a local children's hospital 960m away. The projectile has a muzzle speed of 100m/s and is directed 53.13 degrees above the horizontal. At the same instant, a contingent of concerned physics enthusiasts, knowing full well that children enjoy midair collisions, fire their own mortar at 36.87 degrees above the horizontal. The enthusiasts are 2.00 km away from the street toughs on the opposite side of the hospital, on a 50.0m high hilltop. a.) Find the initial velocity of the enthusiast's projectile if the hospital is to be saved. b.) Calculate the coordinates of the impact of the projectile. Note: use g = 10m/s^2 for the problem. Ignore the dimensions of the hospital.

First I determined the time it took for the street tough's mortar to reach the hospital:
960m = (V * cos53.13)t
960m = (60 m/s)t
t = 16s

Then I tried to figure out what the intitial velocity for the enthusiast's mortar would have to be to travel 1040m (2000-960) in 16s and got:
x=vt
1040m=(v * cos36.87) *16s
1040m = 12.8 * V
81.25 = V

I'm not sure what I'm doing wrong but I'm getting 81.25 m/s and the correct answer is
 
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If you calculate the vertical position of your interception projectile after 16 seconds, you'll see that it does not match the location of the hospital. It would hit the ground much earlier.
The goal is not to hit the hospital at the same time. The goal is to hit the missile in flight before it hits the hospital. Both projectiles have to be at the same place at the same time. Preferably not directly at the hospital.
 
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I tried to set y1 (elderly) = y 2 (enthusiasts) and that didn't work either

(v1*sinθ1)t - 1/2(gt2) = 50 + (v2*sinθ2)t - 1/2(gt2)
100*sin53.13 = 50 + v2 * sin36.87
30 = v2 * sin36.87
50 = v2
 
lynnx95 said:
I tried to set y1 (elderly) = y 2 (enthusiasts) and that didn't work either

(v1*sinθ1)t - 1/2(gt2) = 50 + (v2*sinθ2)t - 1/2(gt2)
100*sin53.13 = 50 + v2 * sin36.87
30 = v2 * sin36.87
50 = v2
What happened to t?
 
Wouldn't it cancel off of both sides? I don't know.

I'm not good at this at all and I have a test in three hours. I don't know how to solve this problem and I really need to be pointed in the right direction for this if someone can please help.
 
lynnx95 said:
Wouldn't it cancel off of both sides?
How? The 50 is not multiplied by t.
If you do it correctly, you get a relation between v2 and t. The horizontal component gives you another relation. Two equations, two unknowns.
 
mfb said:
How? The 50 is not multiplied by t.
If you do it correctly, you get a relation between v2 and t. The horizontal component gives you another relation. Two equations, two unknowns.

So 30t=50+.6v2t?
Not sure what to do with that
 
lynnx95 said:
So 30t=50+.6v2t?
Not sure what to do with that
Write the horizontal equation, as mfb suggests.
Do you not know how to solve simultaneous equations? Standard procedure: if the two unknowns are x and y, and it's x that you want to find, write one equation in the form y=f(x) and use that to replace y in the other equation.
 

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