Calculating Time of Flight for a Vertically Thrown Stone

[jernobyl]

Homework Statement

A stone of mass 3kg is thrown vertically in the air with an initial velocity of 2.8m/s. There are no forces acting on the stone except for that due to gravity and the thrower. Show that the overall time taken for it to return to its point of release is 4/7 seconds.

Homework Equations

v = u + at
v^2 = u^2 + 2as

The Attempt at a Solution

Really all I want to know here is if I'm making more work for myself than is necessary. Taking the motion in two stages:

1. from the throwing until the stone is at momentary rest,
2. from the momentary rest until it reaches the thrower's hand again

1: u = 2.8, v = 0, a = -g

t = (v - t)/a
t = (0 - 2.8)/-9.8
t = 2/7 seconds.

Now, at this point, I REALLY want to believe that I can just multiply that by two for its downward motion and that's it. But I'm in doubt: it's starting with zero velocity, not 2.8m/s velocity like in stage 1. Can I just assume that the time taken for it to return will just be the same again, 2/7 seconds, thus completing the task, or is there more to this?

 

[PhanthomJay]

Homework Statement

A stone of mass 3kg is thrown vertically in the air with an initial velocity of 2.8m/s. There are no forces acting on the stone except for that due to gravity and the thrower. Show that the overall time taken for it to return to its point of release is 4/7 seconds.

Homework Equations

v = u + at
v^2 = u^2 + 2as

The Attempt at a Solution

Really all I want to know here is if I'm making more work for myself than is necessary. Taking the motion in two stages:

1. from the throwing until the stone is at momentary rest,
2. from the momentary rest until it reaches the thrower's hand again

1: u = 2.8, v = 0, a = -g

t = (v - t)/a
t = (0 - 2.8)/-9.8
t = 2/7 seconds.

Now, at this point, I REALLY want to believe that I can just multiply that by two for its downward motion and that's it. But I'm in doubt: it's starting with zero velocity, not 2.8m/s velocity like in stage 1. Can I just assume that the time taken for it to return will just be the same again, 2/7 seconds, thus completing the task, or is there more to this?

No, that's about it. You might want to check your result using s = v_o(t) - 1/2(g)t^2, where s =0, to confirm it. The stone must end up from where it was thrown with the same speed in the opposite direction.

 

[denverdoc]

yes you can, that is the symmetry we so cherish. However this is not always the case so you must carefully examine conditions, for instance if air resistance were an issue, energy is lost throughout, and the descent would never reach the original velocity. But to satisfy whatever lingering doubts it is also possible to take both legs of the journey into acct with the following single eqn:

Yo=Vo*(t)-1/2At^2. now if yo is 0, no overall displacement,
t^2=2*Vo*t/A
t=2*Vo/A =5.6/9.8=4/7 The trick here is in carefully choosing signs, the initial velocity was chosen to be positive and since acceleration doe to gravity the entire time was oppositely directed, it is negative.

 

[jernobyl]

Ah, okay. That's really helped. Thanks, all! :)