Projectile Motion/Dropping an object from the sky

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An eagle flying horizontally at 7.9 m/s drops a fish, which maintains its horizontal speed due to the lack of horizontal acceleration. The discussion highlights that the fish's horizontal speed will never double as it remains constant at 7.9 m/s. For the fish's speed to double, it would need a downward velocity component influenced by gravity, which is not the case here. The calculations presented show that without vertical acceleration, the fish's speed in the x-direction remains unchanged. The problem would be clearer if it involved a downward initial speed, allowing for the effects of gravitational acceleration to be considered.
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An eagle is flying horizontally at 7.9 m/s with a fish in its claws. It accidentally drops the fish.

(a) How much time passes before the fish's speed doubles?

(b) How much additional time would be required for the fish's speed to double again?


Initial velocity in the x direction is 7.9m/s. I assumed final velocity in x direction was 7.9m/s and acceleration was 0. But when I use the equations to find one of the other unknown variables everything cancels out and I get zero. There is an example in the book, but it gives the height the fish is dropped from. I feel like I need to know another variable in order to solve this.
 
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Are you sure you're reading the question right? If the fish is in free fall the only acceleration on it is due to gravity. The speed in the x direction will NEVER double unless there is some acceleration in the x direction.

The distance may double if you assume it can fall long enough, perhaps that is what the question is after.
 
That's what I did. I had Ax=0 m/s^2, Vix= 7.9 m/s, Ay= -9.8 and Viy= 0 m/s and when I plug them into any equation it is always zero
 
Could you show your calculation?
 
I can find either displacement or time. With either one I get 0.

Vf= Vi + at --> 7.9= 7.9 + (0)t
Vf^2= Vi^2 + 2ax --> 7.9^2= 7.9^2 + 2(0)x
 
Amber430 said:
I can find either displacement or time. With either one I get 0.

Vf= Vi + at --> 7.9= 7.9 + (0)t
Vf^2= Vi^2 + 2ax --> 7.9^2= 7.9^2 + 2(0)x

What are you finding the displacement of or what would you be finding the time of?
 
(a) How much time passes before the fish's speed doubles?

(b) How much additional time would be required for the fish's speed to double again?
 
Amber430 said:
(a) How much time passes before the fish's speed doubles?

(b) How much additional time would be required for the fish's speed to double again?

If the eagle is flying completely horizontal at 7.9m/s when it drops the fish then its speed in the horizontal will remain constant; it will never double. The fish could sail through free space for eternity and it's speed will remain 7.9m/s.

The question would make more sense if the 7.9m/s was downwards and the eagle let go. Because then you're finding out how long it takes the acceleration due to gravity to speed up the fish from -7.9m/s to -15.8m/s.
 
What they are asking in this problem is at what time does the speed double?

Speed is the |velocity|.

Hence if Vx is invariant, the magnitude of the velocity vector will be when

Vy is equal to (√3)*Vx ---> Vx² + ((√3)*Vx )² = (1 + 3)*Vx² = 4*Vx² ---> |New V| = 2*|Vx|

So all you need to determine is how long until the Vy is (√3)*Vx
 
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