Calculating Time in Air for a Thrown Ball

[ecthelion4]

Homework Statement

A student standing on the ground throws a ball straight up. The ball leaves the studen'ts hand with a speed of 15 m/s when the hand is 2.0m above the ground. How long is the ball in the air before it hits the ground?

Homework Equations

I'm not sure

The Attempt at a Solution

Ok so I'm guessing I have to imply gravity will have an effect on the ball, I'm just not sure how it comes in the equation with the speed. and since there are no forces working on the x-axis there should be no mention of range. I also checked on the hyper phisics site under free fall but the only relevant topic gives me this equation: Vy=Voy-gt, but that doesn't quite work for me since I can't clear it for t which is the variable I'm looking for. I need help.

 

[radou]

You need an equation which will describe the position of the ball (i.e. the 'y coordinate') with respect to time.

 

[cristo]

You need to use the kinematic equations for motion with constant acceleration, which can be found http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/1DKin/U1L6a.html You know v_i, a and d and want to find t, so pick a suitable equation.

edit: Damn, radou typed faster than me!

 

[ecthelion4]

I guess I'd use this one:

d=Vi*t+1/2gt^2

It's the one that would work in my case since i got speed and g, but what do you say is d? is it the 2m the ball is apart from the x axis?

 

[radou]

I guess I'd use this one:

d=Vi*t+1/2gt^2

It's the one that would work in my case since i got speed and g, but what do you say is d? is it the 2m the ball is apart from the x axis?

You missed a sign in your equation. Think about which one it could be. Further on, you forgot to add another term in your equation. (Hint: initial ________.)

 

[ecthelion4]

Oh I think I get it now

The equation is d=vi*t+1/2*a*t² , so if I substitute the values I know I get this:

2m=15m/s*t+1/2*9.8m/s²*t²

so now i just clear in terms of t right?

 

[radou]

Oh I think I get it now

The equation is d=vi*t+1/2*a*t² , so if I substitute the values I know I get this:

2m=15m/s*t+1/2*9.8m/s²*t²

so now i just clear in terms of t right?

Another hint: in what direction is the initial velocity pointing, and in what direction does gravity act? What does that tell you about the signs of these quantities? Further on, add the initial height to your equation.