Can Physics Save the Hospital from Mortar Fire?

[lynnx95]

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

 

[mfb]

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.

 

[lynnx95]

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

 

[haruspex]

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?

 

[lynnx95]

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.

 

[mfb]

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.

 

[lynnx95]

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

 

[haruspex]

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.