Projectile Puzzle: Which Ball Stays in the Air Longer?

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When two balls are thrown at different angles but reach the same height, they will remain in the air for the same amount of time. The discussion highlights that the vertical component of each ball's velocity must be equivalent for them to achieve the same height. Despite initial confusion regarding the relationship between angle, speed, and time in the air, it is clarified that gravity is the only vertical force acting on both balls. Therefore, both balls will experience the same duration of flight regardless of their launch angles. The conclusion is that both balls must spend the same time in the air.
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Two balls are thrown in the air at different angles, but each reaches the same height. Which ball remains in the air longer?

a the one thrown at the steeper angle
b the one thrown at a shallower angle
c both remain in the air the same time
d need more information

I'm thinking that V must be greater for the lower angle ball. And V = V0 for both.
At first i thought that the time that an object spends in the air is proportional to its height.
However, I'm confused because using t=V0/g results in a greater t for the faster projectile.

Please help. Thank you
 
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Think about the vertical component of each balls velocity. What must they be if the balls reach the same height?
 
Using Y-Y0=Vy0t-1/2gt^2 for the vertical component, and considering that Vy0 will be different for both while the height is fixed, than the one with the greather Vy0 will spend more time in the air. However, given that the only vertical force acting on an object is gravity, they must spend the same amount of time in the air...?
 
Yes both balls must have the same vertical component of velocity. If that is true they must both spend the same time in the air.
 
Ok. So the speed and angle would result in a vertical velocity component that would be equivalent between the two. Thanks you
 
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