Solve Kinematics Problem: 2 Stones Meeting at Height

[marco101]

Homework Statement

You throw a stone straight up with initial V of 24.5m/s and 2.0 s later you throw a second stone straight up with the same initial velocity. The first stone going down will meet the second stone going up. At what height above the point of release do the two stones meet?

Homework Equations

I'm not sure if I'm doing the question correctly. My attempt at it is below.

The Attempt at a Solution

1st Stone
d = Vt
d = 24.5t

2nd Stone
d = 24.5t + .5(9.8)t^2
after 2 seconds, 24.5 x 2 = 49, so d + 49 = 24.5t +.5(9.8)t^2

I then plugged in the d equation from the 1st stone
24.5t + 49 = 24.5t + 4.9t^2
4.9t^2 = 49
t = 3.162 seconds is when the two stones are at the same distance.

Then i assume that you just plug in the t-value in the distance equation for the 1st stone but i keep getting an incorrect answer. Any help please?

 

[tiny-tim]

Welcome to PF!

Hi marco101! Welcome to PF!

… 1st Stone
d = Vt
d = 24.5t

Nope … that's for constant velocity, isn't it?

You need an equation just like for the 2nd stone, but with an adjustment to t.

 

[marco101]

Is my 2nd equation correct then?

Im thinking if it's 2 seconds later, then the equation to the first stone is
d = 24.5 (t+2) + .5(9.8)(t+2)^2

while the 2nd stone stays the same, d + 49 = 24.5t +.5(9.8)t^2 and then set the two equal to each otehr?

 

[tiny-tim]

Hi marco101!

(just got up :zzz: …)

Is my 2nd equation correct then?

Im thinking if it's 2 seconds later, then the equation to the first stone is
d = 24.5 (t+2) + .5(9.8)(t+2)^2

Yup!

while the 2nd stone stays the same, d + 49 = 24.5t +.5(9.8)t^2 and then set the two equal to each otehr?

Nooo … forget the 49 … they both start from d = 0 (one at t = 0, one at t = -2), don't they?