Kinematics, One Dimension, Acceleration, 2 objects problem

[Spoti112]

Homework Statement

Ball A is dropped from the top of a building of height h at the same instant that ball B is thrown vertically upward from the ground. At what height will the balls collide if the collision occurs when the balls are moving in the same direction and the speed of A is 4 times that of B.

V_A0 = 0 - the internal speed of ball A.
h=h; a=g;
t_c - the time when the balls collide.
V_A(tc)=4*V_B(tc)

Homework Equations

[/B]
v = v(initial) +at
x = x(initial) + v(initial)t + (at^2)/2

The Attempt at a Solution

X_A(tc)=X_B(tc)=h_c (height whenn the balls collide)

X_A(tc)= h-(g*(tc)²)/2

X_B(tc)= X_B(tb) - X_B(tb→tc)=
=V_BO - (g*(tb)²)/2 - (g*(tc-tb)²)/2=
=V_BO - (g*(tb)²) - (g*(tc)²)/2 + g*tc*tb
V(tb)=0 ⇒V_B0=g*tb⇒
⇒X_B(tc)=g*tc*tb - (g*(tc)²)/2

X_A(tc)=X_B(tc)= h=g*tc*tb

V_A(tc)=4*V_B(tc) ⇒ -g*tc = -g*(tc-tb) ⇒ 3tc= 4tb →witch is impossible!

 

[TSny]

Hello and welcome to PF!

I think I follow your work up to the following

X_B(tc)= X_B(tb) - X_B(tb→tc)=
=V_BO - (g*(tb)²)/2 - (g*(tc-tb)²)/2=

I'm guessing that tb denotes the time at which B reaches maximum height. So, you are breaking up B's flight into two parts: the upward moving part and the downward moving part. It's not wrong to do this, but there is no need to break it up. The acceleration of B is constant throughout its entire flight. So, you can use the equation X_B = V_B0t - (g/2)t² for the entire interval from t = 0 to t = tc.

The mistake you made is in the second line quoted above where V_BO should be multiplied by a time.

 

[Spoti112]

Hello and welcome to PF!

I think I follow your work up to the following

I'm guessing that tb denotes the time at which B reaches maximum height. So, you are breaking up B's flight into two parts: the upward moving part and the downward moving part. It's not wrong to do this, but there is no need to break it up. The acceleration of B is constant throughout its entire flight. So, you can use the equation X_B = V_B0t - (g/2)t² for the entire interval from t = 0 to t = tc.

The mistake you made is in the second line quoted above where V_BO should be multiplied by a time.

Hello and welcome to PF!

I think I follow your work up to the following

I'm guessing that tb denotes the time at which B reaches maximum height. So, you are breaking up B's flight into two parts: the upward moving part and the downward moving part. It's not wrong to do this, but there is no need to break it up. The acceleration of B is constant throughout its entire flight. So, you can use the equation X_B = V_B0t - (g/2)t² for the entire interval from t = 0 to t = tc.

The mistake you made is in the second line quoted above where V_BO should be multiplied by a time.

thank you very much! I got the correct answer. <3
p.s. sorry that i didn't clarified what tb means

 

[TSny]

OK. Good work.