Calculating Time and Final Velocity of a Ball Thrown into the Air

[Amel]

Homework Statement

The problem: A woman standing on a diving board 21 meters above a pool tosses a ball upwards at a speed of 19 meters/second. The ball goes up and then down, splashing into the water below. How long was the ball in the air? What is the ball’s final velocity?

Homework Equations

v²= v₀²+2a(x-x₀)
X = X₀ +V₀t+(1/2)at²

The Attempt at a Solution

So I figured out the height of the ball by using v²= v₀²+2a(x-x₀) and having v = 0 because that's what it is at the peak and re aranged it so it gives me X. I got 39.25 meters. But I don't know how to get the time the ball is in the air. I assume I have to possibly use X = X₀ +V₀t+(1/2)at² but I am not sure how to get time out of it.

 

[206PiruBlood]

You are correct in that your position function allows you to solve for time. If you take the pool to be at height zero meters, then what is your initial height? You know the acceleration due to gravity and the initial velocity. Put these figures into your equation of motion and you are left with a function of the variable t.

 

[Amel]

Ok so I did t=V₀+/- The square root of ((V₀² + 2y₀g))/g and got 4.94 seconds, The equation came form a problem in the book with a similar problem. But it kind of seems to short of a time. I feel like its not counting the time the ball is traveling up or something.

Anyway is that right? want to know before I do the last part of the equation.

 

[206PiruBlood]

Use the equation $$x(t)= x_0{}+v_0t-\stackrel{1}{2}gt^{2}$$.

My bad, misinterpreted your comment. I just put it in a calculator and got 4.775. Did you use 19m/s as your initial velocity and -9.8 for acceleration? It is much easier to find the time first then the final velocity second.