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
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.
To determine whether the water balloon will hit the fire, overshoot it, or land short of it, we can consider the components of velocity and acceleration.
The angle at which the plane releases the water balloon causes it to have both a horizontal and vertical velocity component. The horizontal velocity component remains unchanged at 60 m/s, as stated in the question. This means that the water balloon will continue to move horizontally at the same speed throughout its flight.
However, due to the tilt of the plane, there will be a downward component of acceleration acting on the water balloon. This acceleration is caused by gravity and is equal to approximately 9.8 m/s².
The combination of the horizontal velocity and the downward acceleration affects the path of the water balloon. The downward acceleration will cause the balloon to fall faster vertically, while the horizontal velocity keeps it moving forward at a constant rate. As a result, the water balloon will follow a parabolic trajectory.
Given that the water balloon is released from the same location and at the same horizontal velocity, the downward acceleration will cause it to land short of the fire. This is because the downward acceleration increases the rate at which it descends vertically, causing it to reach the ground sooner.
Therefore, due to the downward tilt of the plane when the water balloon is released, it will land short of the fire.