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A ball is thrown straight up from the ground with speed v0. At the same instant, a second ball is dropped from rest from a height H, directly above the point where the first ball was thrown upward. There is no air resistance.a) Find the time at which the two balls collide. Express your answer in terms of the variables H, v0, and appropriate constants.b) Find the value of H in terms of v0 and g so that at the instant when the balls collide, the first ball is at the highest point of its motion. Express your answer in terms of the variables v0 and g.

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MechEngineer

Answer:

a)[tex]t = H/v_0[/tex]

b)[tex]H = v_0^2/g[/tex]

Explanation:

Let the first ball throw be the point of reference, we can have following the equation of motion:

1st ball: [tex]h_1 = v_0t - gt^2/2[/tex]

2nd ball: [tex]h_2 = H - gt^2/2[/tex]

a)When the 2 balls collide they are at the same spot at the same time:

[tex]h_1 = h_2[/tex]

[tex]v_0t - gt^2/2 = H - gt^2/2[/tex]

[tex]v_0t = H[/tex]

[tex]t = H/v_0[/tex]

b) The first ball is at its highest point when v = 0. That is

[tex]t = v_0/g[/tex]

After this time, the 2 balls would have traveled through a distance of

[tex]h_1 = v_0t - gt^2/2 = v_0^2/g - v_0^2/2g[/tex]

[tex]h_2 = gt^2/2 = v_0^2/2g[/tex]

Since[tex]H = h_1 + h_2[/tex] we can solve for H

[tex]H = v_0^2/g - v_0^2/2g + v_0^2/2g = v_0^2/g[/tex]

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