Physics, 18.10.2019 20:00 anonymous176

Aphysics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10.0 m/s. the instant before the stone hits the ground below, it is traveling at a speed of 30.0 m/s. if the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10.0 m/s, what is the magnitude of the velocity of the stone just before it hits the ground? ignore any effects of air resistance.

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Answer from: katet2965

vf = 30 m/s : (the magnitude of the velocity of the stone just before it hits the ground)

Explanation:

Because the stone moves with uniformly accelerated movement we apply the following formulas:

vf²=v₀²+2*g*h Formula (1)

Where:

h: displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

g: acceleration due to gravity in m/s²

Free fall of the stone

Data

v₀ = 10 m/s

vf = 30.0 m/s

g = 9,8 m/s²

We replace data in the formula (1) to calculate h:

vf²=v₀²+2*g*h

(30)² = (10)² + (2)(9.8)*h

(30)²- (10)²= (2)(9.8)*h

h =( (30)²- (10)²) /( 2)(9.8)

h = 40.816 m

Semiparabolic movement of the stone

Data

v₀x = 10 m/s

v₀y = 0 m/s

g = 9.8 m/s²

h= 40.816 m

We replace data in the formula (1) to calculate vfy :

vfy² = v₀y² + 2*g*h

vfy² = 0 + (2)(9.8)( 40.816)

$v_{fy}=\sqrt{2*9.8*40.816} = 28.284 \frac{m}{s}$

$v_{f}=\sqrt{v_{ox}^2+v_{fy}^2}=\sqrt{(10)^2+(28.284)^2} = 30\frac{m}{s}$

Answer from: ethanyayger

The magnitude of the velocity of the stone just before it hits the ground is 30 m/s.

The given parameters;

initial vertical velocity of the stone, v_y_0 = 10 m/s

final vertical velocity of the stone, v_y_f = 30 m/s

The height traveled by the stone before it hits the ground is calculated as;

$v_y_f^2 = v_y_0^2 + 2gh\\\\h = \frac{v_y_f^2- v_y_0^2}{2g} \\\\h = \frac{(30)^2 - (10)^2}{2\times 9.8} \\\\h = 40.82 \ m$

If the the stone is projected horizontally with initial velocity of 10 m/s;

the initial vertical velocity = 0

Final vertical velocity of the stone is calculated as follow;

$v_y_f^2 = v_y_0^2 + 2gh\\\\v_y_f^2 = 0 + 2\times 9.8\times 40.82\\\\v_y_f^2 = 800.07\\\\v_y_f = \sqrt{800.07} \\\\v_y_f = 28.28 \ m/s$

The horizontal velocity doesn't change.

the final horizontal velocity, v_x_f = initial horizontal velocity = 10 m/s

The resultant of the final velocity of the stone before it hits the ground;

$v _f= \sqrt{v_x_f^2 + v_y_f^2} \\\\v_f = \sqrt{10^2 + 28.28^2} \\\\v_f= 29.99 \ m/s \approx 30 \ m/s$

Thus, the magnitude of the velocity of the stone just before it hits the ground is 30 m/s.

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Answer from: Quest

Answer: buoyancy forceexplanation: in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

Answer from: Quest

would be false hope this out

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Aphysics student standing on the edge of a cliff throws a stone vertically downward with an initial...