How Far Should an Air Tanker Release Its Load to Hit a Target?

[kle89]

Problem
An air tanker flies horizontally at an altitude of 180 m and a speed of 80 m/s over a forest fire. In
order to hit its target with flame retardant chemicals, how far before it is directly over the target should it
release its load? Neglect air resistance.

Using Projectile motion equations
for the horizontal distance, I set it = to x = (80m/s)(t)
for the vertical distance i used y = y' + vy't - 1/2gt^2 getting y = 180 - (.5)(10 m/s^2)(t^2)

I set them equal to each other

(80m/s)(t) = 180 - (.5)(10 m/s^s)(t^s)
solved for t and got 2 seconds.

Am i doing things right so far or did i mess up somewhere? Thanks in advance?

 

[arunbg]

No, why did you equate x displacement and y displacements ?
What is required is that by the time the chemicals reach the ground, they must have moved horizontally to the exact point of the fire. Find the time taken for the chemicals to reach the ground. Then you can find the tihe distance it travels in that time and thus get the drop distance. Do you follow ?

 

[hage567]

"y = 180 - (.5)(10 m/s^2)(t^2)"

This makes no sense the way it is. You know the total distance it falls, what are you trying to solve for here? (y is zero, since it is at ground level).

"(80m/s)(t) = 180 - (.5)(10 m/s^s)(t^s)"

This doesn't make sense either. You can't equate y to x to solve for t, which is what it looks like you're trying to do. They are not representing the same distance.

What you need to consider is how long it will take the chemicals to fall from the height of 180 m. Once you have that you can figure out what x should be.

 

[kle89]

Thanks for the input.

I used y = y' + vy't - 1/2gt^2
0 = 180m + 0 - 1/2(9.8m/s^2)t^2
4.9t^2 = 180m
t = 6s

So i would plug this into the horizontal velocity x = 80m/s(6s) = 480m

Thanks, is this correct? perhaps i was thinking way to hard earlier.

 

[hage567]

Thanks for the input.

I used y = y' + vy't - 1/2gt^2
0 = 180m + 0 - 1/2(9.8m/s^2)t^2
4.9t^2 = 180m
t = 6s

So i would plug this into the horizontal velocity x = 80m/s(6s) = 480m

Thanks, is this correct? perhaps i was thinking way to hard earlier.

That looks good to me.