Answer:
a) The balls collide at a height = h*(1 - 1/2 g * h/v0²)
b) The maximum value of h is h = v0²/ g
c) The value of h for which the collision occurs at the instant when the first ball is at its highest point is h = v0²/ g (same as b)).
Explanation:
The equations for the position and velocity of an object moving in a straight line with constant acceleration are:
y = y0 + v0*t + 1/2 * a * t²
v = v0 + a * t
where
y = height at time t
y0 = initial height
v0 = initial velocity
a = acceleration (in this case, the acceleration is due to gravity, then, a = g)
t = time
v = velocity at time t
For the first ball that starts on the ground, y0 = 0, then, the equation for the height of the first ball will be:
h₁ = v0 * t - 1/2 g * t² (g is negative because we consider upward as positive and g is directed downward).
For the second ball, that starts at rest, v0 = 0, then, the equation for the height of the second ball will be.
h₂ = h - 1/2 * g * t² (where h is the initial height)
a) When the balls collide, h1 = h2. Then,
h1 = h2
v0 * t - 1/2 g * t² = h - 1/2 * g * t²
v0 * t = h
t = h / v0
Replacing the time t = h/v0 in the equation of any of the balls, we will get the height at which both balls have the same height and collide.
Replacing in the equation of the height of the first ball:
h1 = v0 * h/v0 - 1/2g * h²/v0²
h1 = h - 1/2 g * h²/ v0²
h1 = h*(1 - 1/2 g * h/v0²)
b) Let´s find at which time the first ball reaches its maximum height. At any time after that, the collision will not occur before the first ball falls back. At that time, the speed of the ball is 0. Then,
v = v0 - g * t = 0
t = -v0/-g
This is the maximum time at which the balls can collide before the first ball falls back. We know from a) that the balls collide at t = h/v0. Then:
h/ v0 = -v0/-g
h = v0²/ g
this is the maximum value of h for which a collision occurs before the first ball falls back to the ground. The collision, in this case, will occur at the instant at which the first ball has reached its maximum height.
c) The value of h for which the collision occurs at the instant when the first ball is at its highest point is the maximum value of h for which the collision occurs before the first ball falls back. That was already obtained in b). Then:
h = v0²/ g
In this exercise we have to have knowledge about the horizontal launch, so we have to use the known formulas to find that:
a) [tex]h*(1 - 1/2 g * h/v_0^2)[/tex]
b)[tex]h = v_0^2/ g[/tex]
c)[tex]h = v_0^2/ g[/tex]
So we have to remember some famous equations like the position and velocity of an object moving in a constant, like this:
[tex]y = y_0 + v_0*t + 1/2 * a * t^2\\v = v_0 + a * t[/tex]
where:
a) When the balls collide, h1 = h2. Then,
[tex]h_1 = h_2\\v_0 * t - 1/2 g * t^2 = h - 1/2 * g * t^2\\v_0 * t = h\\t = h / v_0[/tex]
Replacing in the equation of the height of the first ball:
[tex]h_1 = v_0 * h/v_0 - 1/2g * h^2/v_0^2\\h_1 = h - 1/2 g * h^2/ v_0^2\\h_1 = h*(1 - 1/2 g * h/v_0^2)[/tex]
b) that the balls collide at t = h/v0. Then:
[tex]h/ v_0 = -v_0/-g\\h = v_0^2/ g[/tex]
c) The value of h for which the collision occurs at the instant when the first ball is at its highest point is the maximum value,Then:
[tex]h = v_0^2/ g[/tex]
See more about velocity at brainly.com/question/862972
