Calculating Velocity of a Ball Projected Upwards

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A ball is projected upwards from a 25 m high building and takes 3 seconds to return to the edge on its way down. The initial velocity can be calculated using kinematic equations, specifically s = so + vot + 1/2gt² and v = vo + at. It's clarified that the time to reach the peak of the arc does not need to be assumed as 1.5 seconds, although it is correct that it takes half the total time to return to the launch height. The key variables needed for the calculation are the initial height and the acceleration due to gravity. Using the kinematic equations will provide the necessary solution for 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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