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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. When the balls collide, they are moving in opposite directions, and the speed of A is twice the speed of B. At what height does the collision occur?

Respuesta :

Answer:

(2/3) times height collision occur

Explanation:

for ball A

from the kinematic equation

the distance of ball A is

[tex]x_A = v_0 t + \frac{1}{2} at^2[/tex]

[tex]v_0* t [/tex]= velocity ( time ) = distance

since ball is at height, the above equation changes as

[tex]x_A = H - \frac{1}{2} gt^2[/tex]

for ball B

[tex]xB = v0 t - \frac{1}{2} gt^2[/tex]

the condition of collision is

xA = xB

[tex]vA = - 2vB[/tex] (given)

from the kinematic equation

the speed of the ball A is

[tex]v_A = u- gt[/tex]

since initial speed of the ball A is zero

, so

[tex]v_A = -gt[/tex]

the speed of the ball B is

[tex]v_B = v_0 - gt[/tex]

since[tex]v_A = - 2v_B[/tex]

   [tex]-gt = -2 ( v_0 - gt)[/tex]

[tex]-gt = -2 v_0 +2gt[/tex]

[tex]3gt =2 v_0[/tex]

[tex]t = \frac{2v_0}{3g}[/tex]

since [tex]x_A = x_B[/tex]

[tex]H -  \frac{1}{2} gt^2= v_0 t - \frac{1}{2} gt^2  [/tex]

[tex]H = v_0 t[/tex]

[tex]= v_0 (2v_0/3g)[/tex]

[tex]= \frac{2 v_0^2}{ 3g}[/tex]

[tex]x_A = 2 (\frac{v_0^2}{ 3g})- \frac{1}{2} gt^2[/tex]

  [tex]= 4 \frac{v_0^2}{9g} = (2/3) H[/tex]

so, (2/3) times height collision occur

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