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3 years ago

Four identical balls are thrown from the top of a cliff, each with the same speed. The

Physics

1 answer:

Jlenok [28]3 years ago

6 0

Answer:

the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal

Explanation:

The kinetic energy, K.E. = (1/2) × m × v²

The velocity of the ball, v = u × sin(θ)

Where;

u = The initial velocity of the ball

θ = The reference angle

1) For the ball thrown straight up, we have;

θ = 90°

∴ v = u

The final velocity of the ball as it strikes the ground is v₂ = u² + 2gh

Where;

h = The height of the cliff

∴ K.E. = (1/2) × m × (u² + 2gh)²

2) For the second ball thrown 30° to the horizontal, we have;

K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²

3) For the third ball thrown at 30° below the horizontal, we have;

K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²

4) For the fourth ball thrown straight down, we have;

K.E. = (1/2) × m × (u² + 2gh)²

Therefore, as the ball strike the ground, the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal

Learn more about object kinetic energy of object if free fall here;

brainly.com/question/14872097

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Vi=0 (because it was dropped, not thrown)
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b) How far did the brick fall after 4s?
D=?
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t=4s
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3 years ago

An object is dropped from a height H. During the final second of its fall, it traverses a distance of 53.2 m. What was H? An obj

Serhud [2]

Answer:

H = 171.90 m

Explanation:

given data

distance = 53.2 m

height = H

to find out

height H

solution

we know height is here H = $\frac{1}{2} gt^2$ ......................1

here t is time and a is acceleration

we find t first

we know during time (t -1) s , it fall distance (H - 53.2) m

so equation of distance

$H - 53.2 = \frac{1}{2} g (t-1)^2$

$H - 53.2 = \frac{1}{2} g (t^2-2t+1)$

$H - 53.2 = \frac{1}{2} gt^2-gt+\frac{1}{2} g$ ................2

now subtract equation 2 from equation 1 so we get

$H - (H - 53.2) =\frac{1}{2} gt^2- (\frac{1}{2} gt^2-gt+\frac{1}{2} g)$

53.2 = gt - $\frac{1}{2} g$

53.2 = 9.81 t - $\frac{1}{2} 9.8$

t = 5.92 s

so from equation 1

H = $\frac{1}{2} (9.81)5.92^2$

H = 171.90 m

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3 years ago

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3 years ago

Read 2 more answers

Suppose you are navigating a spacecraft far from other objects. The mass of the spacecraft is 2.0 104 kg (about 20 tons). The ro

elena-14-01-66 [18.8K]

Answer:

r = <2.640 10⁶, 1.01 10⁸, 0> m

Explanation:

For this exercise we are going to solve it for each direction separately,

we locate a fixed reference frame in space at the height of the rocket, such that the position of the rocket is

r₀ = <4, 4, 0> 10³ m

X axis

the initial velocity of the ship on this axis is v₀ₓ = 0, when it passes through the point x₀ = 4 km it ignites the rockets, experiencing a force of Fₓ = 7.0 10⁵ N for 21 s and the rockets turn off

They ask where it is after one hour t = 1 h = 3600 s

Let's apply Newton's second law

Fₓ = m aₓ

aₓ = Fₓ / m

aₓ = 7.0 10⁵/2 10⁴

aₓ = 3.5 10¹ m / s

Let's use kinematics to find the distance

for the first t₁ = 21 s the movement is accelerated

x₁ = x₀ + v₀ t₁ + ½ aₓ t₁²

x₁ = x₀ + ½ aₓ t₁²

x₁ = 4000 + ½ 35 21² = 4000 + 7717,.5

x₁ = 11717.5 m

this instant has a speed of

vₓ = v₀ₓ + aₓ t

vₓ = aₓ t ₁

vₓ = 35 21

vₓ = 735 m / s

the rest of the time there is no acceleration so it is uniform motion at this speed

t₂ = 3600 - 21

t₂ = 3576 s

vₓ = x₂ / t₂

x₂ = vₓ t₂

x₂ = 735 3576

x₂ = 2,628 10⁶ m

the total distance traveled in this direction is

x_total = x₁ + x₂

x_total = 11717.5 + 2.628 10⁶

x_total = 2,640 10⁶ m

Y axis

on this axis it is in the initial position of y₀ = 4000 m, with an initial velocity of $v_{oy}$ = 28 10³ m / s and there is no force on this axis F_{y} = 0

The movement in this axis is uniform,

v_{y} = y / t

y = v_{y} t

y = 28 10³/3600

y = 1.01 10⁸ m

the total distance is

y_total = y₀ + y

y_total = 4000 + 1.01 10⁸

y_total = 1.01 10⁸ m

Z axis

the initial position is z₀ = 0, with an initial velocity of v₀ = 0 and in this axis there is no force F_{z} = 0

the movement is uniform

z = 0

the final position of the rocket after 1 h is

r = <2.640 10⁶, 1.01 10⁸, 0> m

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3 years ago