Finding total flight time of a projectile. Please help

[Ilovepysics]

Homework Statement

A baseball thrown vertically with a speed of 3m/s. Find the total time that the ball has been in flight when it had traveled 50cm.

Homework Equations

vf=vi + at
d= 1/2at^2

The Attempt at a Solution

Found t.
vf=vi +at
vf=3
vi=0
a=~10
Found t= .3 secs

Found d
d=1/2 + at^2
d=4.5m

Added 4.5m to .5m =5m

Plugged in total distance 5m into d=1/2 + at^2
Found t= 1

Added 1+.3 to get total time of 1.3 secs.

The answer in the manual is 0.398 sec.
I am way off can someone please help. Thanks!

 

[grzz]

d= 1/2at^2

above equation is not always correct.

 

[Ilovepysics]

grzz what do you mean?

Is there another equation I should be using? or perhaps another approach. I am not sure where I went wrong.

 

[grzz]

The equation that I quoted from your post is only correct to use if the initial speed is 0. This is not the case in your problem.

 

[Ilovepysics]

I also tried using d=vit + 1/2at^2 plugging in 3m/s for vi but I also calculated a wrong answer for t. So I am not sure what equation to use.

 

[TaxOnFear]

Use the constant acceleration equations. One in particular will allow you to find the final velocity at S = 0.5m, then you can use that to calculate other values e.g time.

 

[Ilovepysics]

TaxOnFear-
I tried using vf^2 =vi^2 + 2ad to find vf. I got sqroot of 19 for vf and then plugged it into vf=vi+ at to solve for t I got t=.135 sec... still the wrong answer according to the answer manual.

 

[physicsvalk]

TaxOnFear-
I tried using vf^2 =vi^2 + 2ad to find vf. I got sqroot of 19 for vf and then plugged it into vf=vi+ at to solve for t I got t=.135 sec... still the wrong answer according to the answer manual.

You are assuming the projectile has not reached the peak point by this statement. Try breaking the problem down into two sections: one where the projectile is launched up, and the other where the projectile is falling down. Most of the numbers will look the same, but the time in the air will add up.

 

[Ilovepysics]

I am really confused. I am starting to think there is an error in the answer manual. I have tried all motion equations possible. I assumed it reached peak and used vf=3m/s -- that didnt work. I tried vi=3m/-- that didnt work. I tried d= 1/2at^2 that didnt work either.

Was my initial attempt at a solution (above) completely wrong?? Can someone tell me where I am going wrong. I've tried looking at it from all angles..please help

 

[grzz]

Better follow the advice of 'physicsvalk'.

The 50cm may mean the total distance going up AND some distance going down.

Hence try to find the MAXIMUM height it can travel. Of course in finding this you have to put the final velocity equal to 0.

Then find the time to reach this max height and continue from there.