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3 years ago

A stone is thrown vertically upwards with a speed of 30.0 m/s.

Physics

1 answer:

matrenka [14]3 years ago

6 0

Y₀ = initial position of the stone at the time of launch = 0 m

Y = final position of stone = 20.0 meters

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = ?

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

v² = 30² + 2 (- 9.8) (20 - 0)

v = 22.5 m/s

Y₀ = initial position of the stone at the time of launch = 0 m

Y = maximum height gained

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = 0 m/s

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

0² = 30² + 2 (- 9.8) (Y - 0)

Y = 46 m

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3 years ago

A rock is thrown horizontally out of a window with a velocity of 20.0 m/s. If the window is 8.50m above the ground, how far

vladimir1956 [14]

The distance from the base of the building the rock will land is 26.4 m

<h3>Data obtained from the question </h3>

Horizontal velocity (u) = 20 m/s
Height (h) = 8.50 m
Distance (s) =?

<h3>Determination of the time to reach the ground </h3>

Height (h) = 8.50 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?

h = ½gt²

8.5 = ½ × 9.8 × t²

8.5 = 4.9 × t²

Divide both side by 4.9

t² = 8.5 / 4.9

Take the square root of both side

t = √(8.5 / 4.9)

t = 1.32 s

<h3>How to determine the distance </h3>

Horizontal velocity (u) = 20 m/s
Time (t) = 1.32 s
Distance (s) =?

s = ut

s = 20 × 1.32

s = 26.4 m

Learn more about motion under gravity:

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3 years ago

The space shuttle releases a satellite into a circular orbit 630 km above the Earth.

solniwko [45]

Answer:

7,539 m/s

Explanation:

Let's use this equation to find the gravitational acceleration of this space shuttle:

$\displaystyle g=\frac{GM}{r^2}$

We know that G is the gravitational constant: 6.67 * 10^(-11) Nm²/kg².

M is the mass of the planet, which is Earth in this case: 5.972 * 10^24 kg.

r is the distance from the center of Earth to the space shuttle: radius of Earth (6.3781 * 10^6 m) + distance above the Earth (630 km → 630,000 m).

Plug these values into the equation:

$\displaystyle g=\frac{(6.67\cdot 10^-^1^1 \ Nm^2kg^-^2)(5.972\cdot 10^2^4 \ kg)}{[(6.3781\cdot 10^6 \ m)+(630000 \ m)]^2}$

Remove units to make the equation easier to read.

$\displaystyle g=\frac{(6.67\cdot 10^-^1^1 )(5.972\cdot 10^2^4 )}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Multiply the numerator out.

$\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Add the terms in the denominator.

$\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(7008100)]^2}$

Simplify this equation.

$\displaystyle g=8.11045189 \ \frac{m}{s^2}$

The acceleration due to gravity g = 8.11045189 m/s². Now we use the equation for acceleration for an object in circular motion which contains v and r.

$\displaystyle a = \frac{v^2}{r}$

a = g, v is the velocity that the space shuttle should be moving (what we are trying to solve for), and r is the radius we had in the previous equation when solving for g.

Plug these values into the equation and solve for v.