Physics, 11.07.2019 19:30 liltonio

Aball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the instant the first ball is released. determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground. (use h and g as appropriate in your equation.)

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Answer from: makaileep1623

The initial velocity of the second ball is $\sqrt{hg}$ 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.

$x_1 = h + \dfrac {1}{2}(-g)t^2$

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.

$x_2 = vt +\dfrac {1}{2} (-g)t^2$

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,

$x_1 = x_2 = \dfrac {h}{2}$

Substituting the value of x1 in the above equation, we get time t as given below.

$\dfrac {h}{2} = h - \dfrac {1}{2} gt^2$

$\dfrac {1}{2}gt^2 = h - \dfrac {h}{2}$

$\dfrac {1}{2}gt^2 = \dfrac {h}{2}$

$t = \sqrt{\dfrac{h}{g}}$

Substituting the value of t and x2 in the above equation, we get the initial velocity v of the second ball.

$\dfrac {h} {2} = v\sqrt{\dfrac {h}{g}} - \dfrac {1}{2} g(\sqrt{\dfrac {h}{g}} )^2$

$\dfrac {h}{2} + \dfrac {1}{2}g (\dfrac {h}{g}) = v \sqrt{\dfrac {h}{g}}$

$\dfrac {h}{2} + \dfrac {h}{2} = v\sqrt{\dfrac {h}{g}}$

$v = \sqrt{hg}$

Hence we can conclude that the initial velocity of the second ball is $\sqrt{hg}$ m/s.

To know more about the velocity, follow the link given below.

link.

Answer from: ciaralamont2

The position of the first ball is

$y_1=h-\dfrac g2t^2$

while the position of the second ball, thrown with initial velocity , is

$y_2=vt-\dfrac g2t^2$

The time it takes for the first ball to reach the halfway point satisfies

$\dfrac h2=h-\dfrac g2t^2$

$\implies\dfrac h2=\dfrac g2t^2$

$\implies t=\sqrt{\dfrac hg}$

We want the second ball to reach the same height at the same time, so that

$\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2$

$\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)$

$\implies h=v\sqrt{\dfrac hg}$

$\implies v=\sqrt{hg}$

Answer from: Quest

step-by-step explanation:

10 - 2.8 = 7.2

2 2/5 = 2.4

7.2 / 2.4 = 3

Answer from: Quest

adding a(n) to an electric circuit would increase the magnetic field around a wire;

b. solenoid

a solenoid is a type of electromagnet whose purpose is to generate a controlled magnetic field.

if the purpose of the solenoid is instead to impede changes in the electric current, a solenoid can be more specifically classified as an inductor rather than an electromagnet.

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