Ball A is dropped from the top of a building at the same instant that ball B is thrown vertically upward from

Question

Ball A is dropped from the top of a building 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 fraction of the height of the building does the collision occur.

Answer 1

Ball A is dropped from the top of a building 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 fraction of the height of the building does the collision occur.
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ans:
here is the formula I know

when a ball is dropped from the top of building of height h
and at the same time another ball is projected vertically
upwards from the ground with an initial velocity u , they
will meet after a time interval of t = h/u-------(1)
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But, in the above problem , u said that they will collide in
opposite directions, but, there is no possibility to collide in
opposite directions. I think u entered data wrongly.

Answer 2

Very cool problem.

You need to use conservation of energy and momentum. I am assuming ball A and B have the same mass.

The speed of ball A at collision is gotten by solving:

mgH = 1/2 * m * v_a ^ 2 + mgh

where H is the height of the building and h is the height above the ground where the collision occurs.

Solving for v_a gives:
-sqrt(2g(H - h)) (negative, since ball A is moving downwards).

Similarly, ball B's speed upon impact is:

1/2 * m * v_o ^ 2 = 1/2 * m * v_b ^ 2 + mgh

where v_o is the initial speed of B.

Solving for v_b yields:

+sqrt(v_o ^ 2 - 2gh) (positive, since B is moving upwards).

Now conservation of momentum:

mv_b + mv_a = -mv + m*2v

where v is the speed of ball B after collision. Note that now A is moving up and B is moving down.

Plug everything in:

sqrt(v_o ^ 2 - 2gh) - sqrt(2g(H - h)) = mv

You can here solve for h in terms of H. You should get h = c*H, where c is some fraction with 0 < c < 1.

Something tells me you have more information which is not given here....

Answer 3

2/3

Proof:

H – height of the building
h – height of the collision
t – time of the collision
u – initial velocity

H – h = (g*t^2)/2

h = u*t - (g*t^2)/2

=> h/H = 1 – (g*t)/(2*u)

When the balls collide, the speed of A is twice the speed of B:

g * t = 2 * (u – g * t)

=> t = (2*u)/(3*g)

h/H = 1 – (g*t)/(2*u) = 1 – (g*(2*u)/(3*g))/(2*u)) = 1 - 1/3 = 2/3 q.e.d.

Answer 4

In time t, the ball A has a speed Va = gt.
At the same time the ball B has a speed Vb = U – gt.

Vb + Va = U, the speed with which the ball B is thrown up.

If Wa and Wb are the speeds after collision, Wa = 2 Wb.

Assuming the balls has equal masses, by conservation of momentum,

Vb – Va = Wa – Wb = Wb.

By conservation of energy,

[Vb^2 + Va^2] = Wa^2 + Wb^2 = 5 Wb^2.

[Vb^2 + Va^2] = 5 Wb^2.

[Vb^2 + Va^2] = [Vb – Va] ^2 + 2 Va Vb

5 Wb^2 = W b^2 +2 Va Vb

2 Wb^2 = Va Vb

[Vb +Va]^2 = [Vb – Va] ^2 + 4 Va b

U^2 = Wb^2 + 8 Wb^2 = 9 Wb^2

U = 3 Wb

Vb + Va = U = 3 Wb

Vb – Va = = Wb.

Vb = 2Wb and Va = Wa.
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Va^2 = 2gH1 or Wa^2 = 2gH1

U^2 – Vb^2 = 2gH2 or 9 Wb^2 – 4Wb^2 = 5 Wb^2 = 2gH2

H1 /H2 = 1/5

Answer 5

permit t be the time they must meet. then the dropped ball has the cost vA = g * t jointly as the growing to be ball has the cost vB = a million/2 vA = a million/2 gt Ball A fell f = a million/2 g t^2 Ball B rose r = a million/4 gt^2 the heigth is h = r + f = 3/4 gt^2 = 3 * r So h = 3 * r ---> r = a million/3 * h

Answer 6

It will occur at 3/4h. If you were to take the time to pick set values, and then graph both x-t graphs, they would intersect at 3/4 the height.