Respuesta :
The position of the first ball is
[tex]y_1=h-\dfrac g2t^2[/tex]
while the position of the second ball, thrown with initial velocity [tex]v[/tex], is
[tex]y_2=vt-\dfrac g2t^2[/tex]
The time it takes for the first ball to reach the halfway point satisfies
[tex]\dfrac h2=h-\dfrac g2t^2[/tex]
[tex]\implies\dfrac h2=\dfrac g2t^2[/tex]
[tex]\implies t=\sqrt{\dfrac hg}[/tex]
We want the second ball to reach the same height at the same time, so that
[tex]\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2[/tex]
[tex]\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)[/tex]
[tex]\implies h=v\sqrt{\dfrac hg}[/tex]
[tex]\implies v=\sqrt{hg}[/tex]
The initial velocity of the second ball is [tex]\sqrt{hg}[/tex] m/s.
How do you calculate the speed of the second ball?
Given that a ball is dropped from rest from a height h above the ground. This ball has acceleration equal to gravitational acceleration. The position of the first ball for the time t is given below.
[tex]x_1 = h + \dfrac {1}{2}(-g)t^2[/tex]
Another ball is thrown vertically upwards from the ground at the instant the first ball is released. Let consider that the initial velocity of the second ball is v and acceleration is equal to gravitational acceleration. The position of the second ball for the time t is given below.
[tex]x_2 = vt +\dfrac {1}{2} (-g)t^2[/tex]
If the two balls are to meet at a height h/2 above the ground. Their position will be the same at the meeting point. It means that,
[tex]x_1 = x_2 = \dfrac {h}{2}[/tex]
Substituting the value of x1 in the above equation, we get time t as given below.
[tex]\dfrac {h}{2} = h - \dfrac {1}{2} gt^2[/tex]
[tex]\dfrac {1}{2}gt^2 = h - \dfrac {h}{2}[/tex]
[tex]\dfrac {1}{2}gt^2 = \dfrac {h}{2}[/tex]
[tex]t = \sqrt{\dfrac{h}{g}}[/tex]
Substituting the value of t and x2 in the above equation, we get the initial velocity v of the second ball.
[tex]\dfrac {h} {2} = v\sqrt{\dfrac {h}{g}} - \dfrac {1}{2} g(\sqrt{\dfrac {h}{g}} )^2[/tex]
[tex]\dfrac {h}{2} + \dfrac {1}{2}g (\dfrac {h}{g}) = v \sqrt{\dfrac {h}{g}}[/tex]
[tex]\dfrac {h}{2} + \dfrac {h}{2} = v\sqrt{\dfrac {h}{g}}[/tex]
[tex]v = \sqrt{hg}[/tex]
Hence we can conclude that the initial velocity of the second ball is [tex]\sqrt{hg}[/tex] m/s.
To know more about the velocity, follow the link given below.
https://brainly.com/question/9163788.