a water bomber flying with a horizontal speed of 85m/s at a height of 3000 m drops a load on a fire below. how far in front of the target fire should the load be released?

h=1/2 at^2

t^2=(2*3000)/9.8
t^2=612.244898
t=24.74358297sec
d=vt
d=85m/s*24.74358297s
d=2103meters

how long does it take to fall 3000m?

h=1/2 a t^2...
t=sqrt(2*3000/9.8)

distance in front=85m/s*timeabove

Lord, I just told you how:

distanceinfront=85*sqrt(2*3000/9.8)

where do you find the answer key?

To determine how far in front of the target fire the load should be released, we need to consider the horizontal distance traveled by the water bomber while the load is falling.

First, let's calculate the time it takes for the load to fall from a height of 3000 m. We can use the formula:

h = (1/2) * g * t^2

where:
h is the height (3000 m),
g is the acceleration due to gravity (9.8 m/s^2),
and t is the time.

To find the time t, we rearrange the formula:

t = sqrt((2h) / g)

t = sqrt((2 * 3000) / 9.8)
t ≈ 24.5 seconds

Next, we calculate the horizontal distance traveled by the water bomber during this time. We use the equation:

d = v * t

where:
d is the horizontal distance,
v is the horizontal speed of the water bomber (85 m/s),
and t is the time (24.5 seconds).

d = 85 * 24.5
d ≈ 2072.5 meters

Therefore, the load should be released approximately 2072.5 meters in front of the target fire.

it's asking for meters. the answer is 2103m just don't know how to get there.