Calculating Velocity of a Ball Projected Upwards

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SUMMARY

The discussion revolves around calculating the initial velocity of a ball projected vertically upwards from a height of 25 meters, given that it takes 3 seconds to return to the edge of the building. Using the kinematic equation s = so + vot + 1/2gt², where g is 10 m/s², participants clarify that the time to reach the peak of the projectile's arc is indeed 1.5 seconds. The solution requires understanding the initial conditions and applying the kinematic equations correctly to find the initial velocity.

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Homework Statement


A ball is projected vertically upwards from the top of a building that is 25 m high. After a time of 3 s it passes the edge of the building on its way down to the ground. In this question, g=10ms-2.

a) What is it's initial velocity?

Homework Equations



s=so + vot + 1/2gt2

v= vo + at

The Attempt at a Solution



Basically, all I want to know is this: Does the question want me to assume that it reaches the top of the projectile 'arc' where velocity in the vertical direction is zero, when the time is 1.5 seconds (half of 3 seconds). Or is there a mathematical way of showing it?

I'd be fine if I knew one more variable, but I know very few variables...

Thanks in advance.
 
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It's not necessary to assume anything abut the time that it reaches the top of its arc (although you are correct that it will take half the time it takes to go from launch height to again be at launch height at the edge of the building).

The first kinematic equation that you wrote is sufficient for your purpose. What's so? What's s when the ball passes the edge?
 

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