Another One Dimensional Motion Problem

[pointintime]

Homework Statement

A rock is thrown vertically upward with a speed of 12.0 s^-1 m. Exactly 1.00 s later, a ball is thrown up vertically along the same path with a speed of 20.0 s^-1 m.

(a) At what time will they strike each other?

(b) At what height will the collision occur?

(c) Answer (a) and (b) assuming that the rder is reversed: the ball is trown 1.00 s before thr ock.

Homework Equations

X = Xo + Vot + 2^-1 a t^2

The Attempt at a Solution

I have no idea how to even start this problem
were the 1 subscripts are the first object that went off and the twos are the second object
t1 = t2 - 1.00s

plugged in

Xo + Vo1 t1 + 2^-1 a t1^2

Xo + Vo1 (t2 - 1.00 s) + 2^-1 a (t2 - 1.00)^2

solved for t2 found 1 second...

 

[pointintime]

if i consider there velocities to be positive do i have to make there accelerations negative?

 

[pointintime]

do they collide at 2.152 s + 1.00 s ?

not sure if that's right...

 

[tiny-tim]

Hi pointintime!

(try using the X² and X₂ tags just above the Reply box )

if i consider there velocities to be positive do i have to make there accelerations negative?

If you consider distance upward to be positive, then yes, you have to make the accelerations negative.

(and if you want us to check your result, do show the whole calculation! )

 

[pointintime]

how do you recommend i solve it

 

[tiny-tim]

how do you recommend i solve it

The method you're using looks fine …

what's worrying you about it? ​

 

[pointintime]

X1 = X2
sense they collide

t1 = t2
point at which the collide

t1 = t2 - 1.00 s

X1 = Xo + Vo t + 2^-1 a t^2

X1 = (12.0 s^-1 m)t1 + 2^-1 (9.80 s^-2 m)t1^2

sense
t1 = t2 - 1.00 s

X1 = (12.0 s^-1 m)(t2 - 1.00 s) + 2^-1 (9.80 s^-2 m)(t2 - 1.00 s)^2
X1 = (12. 0 s^-1 m)t2 - 12.0 m + 2^-1 (9.80 s^-2 m)(t2^2 + 1.00 s^2 - 2 t2 1.00s)
X1 = (12. 0 s^-1 m)t2 - 12.0 m + (4.90 s^-2 m)(t2^2 + 1.00 s^2 - t2 2.00s)
X1 = (12. 0 s^-1 m)t2 - 12.0 m + (4.90 s^-2 m)t2^2 + 4.90 m - (9.80 s^-1 m)t2
X1 = (4.90 s^-2 m)t2^2 + (2.2 s^-1 m) - 7.1 m

sense X1 = X2

(4.90 s^-2 m)t2^2 + (2.2 s^-1 m) - 7.1 m = Xo + Vo t2 + 2^-1 a t2^2
(4.90 s^-2 m)t2^2 + (2.2 s^-1 m) - 7.1 m = (20.0 s^-1 m) t2 + 2^-1 (9.80 s^-2 m) t2^2
(4.90 s^-2 m)t2^2 + (2.2 s^-1 m) - 7.1 m - (20.0 s^-1 m) t2 -(4.90 s^-2 m) t2^2
(17.8 s^-1 m)t2 - 7.1 m

what now?
let me guess I did it wrong

(17.8 s^-1 m)t2 - 7.1 m = 0
(17.8 s^-1 m)t2 = 7.1 m
t2 = .3989 s

what does that mean?

 

[pointintime]

can someone check my work i probably did it wrong and if i didn't i don't know what to do from there

 

[tiny-tim]

can someone check my work i probably did it wrong and if i didn't i don't know what to do from there

Well, it's bit difficult to read (please use the X² tag in future), but the method looks ok,

except (i thought you agreed on this) the acceleration needs to be negative.

 

[pointintime]

It does have to be negative?
If I consider it to be positive does it miss anything up

 

[tiny-tim]

It does have to be negative?
If I consider it to be positive does it miss anything up

Yeees! …

if your 12 and 20 are positive, then your 9.8 must be negative (or vice versa).

what's up for velocity must be up for acceleration!​