If the plane is descending when it releases the water, it starts out with some nonzero downward velocity. So it will hit the ground sooner. Its height
y = h+vt - 1/2 gt^2
If v < 0, y decreases faster.
There are roughly 8000 wildfires in Canada each year. To help fight the fires, planes are used to drop water and fire retardants on the flames. One such plane flies horizontally over a fire at a speed of 60 m/s and drops a giant water balloon to help extinguish the fire. It flies at a height of 200 m.
The pilot accidentally angles his plane slightly downward when releasing his water balloon, with the nose lower than the tail. He releases the water balloon from the exact same location as before. Will the balloon still hit the fire, overshoot it, or land short of it? You don’t need to do any explicit calculations, but use what you know about velocity and acceleration to justify your answer.
I'm really unsure how to go about this. I'm sure that the acceleration would stay the same since its just the gravity of the earth and I can't imagine how tilting it forward would increase velocity or decrease it
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