Ask question

Ask question

All categories

3 years ago

9

A ball is thrown vertically upward from the edge of a bridge 22.0 m high with an initial speed of 16.0 m/s. The ball falls all t

he way down and strikes the water below. Determine the magnitude of the velocity of the ball just before it strike the water.

1 answer:

KonstantinChe [14]3 years ago

3 0

To solve this problem we will apply the concepts related to the kinematic equations of motion. We will start calculating the maximum height with the given speed, and once the total height of fall is obtained, we will proceed to calculate with the same formula and the new height, the speed of fall.

The expression to find the change in velocity and the height is,

$v_f^2-v_0^2 = -2gh$

Replacing,

$0^2-16^2 = -2(9.8)h$

$h = 13.0612m$

Thus the total height reached by the ball is

H = 22m+13.0612m

H = 35.0612m

Now calculate the velocity while dropping down from the maximum height as follows

$v_f^2-v_0^2 = 2gh$

Substituting the new height,

$v_f^2 - 0^2 = 2(9.8)(35.0612)$

$v = \sqrt{687.2}$

$v = 26.2145m/s$

You might be interested in

How many lenses with different focal lengths can be obtained by combining two surfaces whose radii of curvature are 4.00 cm and

Dovator [93]

Answer:

The lenses with different focal length are four.

Explanation:

Given that,

Radius of curvature R₁= 4

Radius of curvature R₂ = 8

We know ,

Refractive index of glass = 1.6

When, R₁= 4, R₂ = 8

We need to calculate the focal length of the lens

Using formula of focal length

$\dfrac{1}{f}=(n-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})$

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}+\dfrac{1}{8})$

$\dfrac{1}{f}=\dfrac{9}{40}$

$f=4.44\ cm$

When , R₁= -4, R₂ = 8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}+\dfrac{1}{8})$

$\dfrac{1}{f}=-\dfrac{3}{40}$

$f=-13.33\ cm$

When , R₁= 4, R₂ = -8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}-\dfrac{1}{8})$

$\dfrac{1}{f}=\dfrac{3}{40}$

$f=13.33\ cm$

When , R₁= -4, R₂ = -8

Put the value into the formula

$\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}-\dfrac{1}{8})$

$\dfrac{1}{f}=-\dfrac{9}{40}$

$f=-4.44\ cm$

Hence, The lenses with different focal length are four.

8 0

2 years ago

lightning is actually an enormous display of the concept of A.chemical heating B.solar energy C.magnetism D.grounding

allsm [11]

Answer:

D) Grounding

Explanation:

The potential difference between cloud and ground leads to ionization of the atmosphere and resulting conduction through the air often to ground (although it can be between clouds at different potentials. I would say grounding, like the spark when you touch a hot battery terminal to ground on a car.

5 0

2 years ago

You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in

FinnZ [79.3K]

Explanation:

It is given that,

The volume of a right circular cylindrical, $V=108\ cm^3$

We know that the volume of the cylinder is given by :

$V=\pi r^2 h$

$108=\pi r^2 h$

$h=\dfrac{108}{\pi r^2}$ ............(1)

The upper area is given by :

$A=32r^2+2\pi rh$

$A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}$

$A=32r^2+\dfrac{216}{r}$

For maximum area, differentiate above equation wrt r such that, we get :

$\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}$

$64r-\dfrac{216}{r^2}=0$

$r^3=\dfrac{216}{64}$

r = 1.83 m

Dividing equation (1) with r such that,

$\dfrac{h}{r}=\dfrac{108}{\pi r}$

$\dfrac{h}{r}=\dfrac{108}{\pi 1.83}$

$\dfrac{h}{r}=59 \pi$

Hence, this is the required solution.

8 0

2 years ago

The quadriceps muscles pull on the patella simultaneously. Below are the forces from each

Nostrana [21]

Based on the calculation of the resultant of vector forces:

the resultant force due to the quadriceps is 1795 N
the resultant force due to the quadriceps is 1975 N. Training and strengthening the vastus medialis results in a greater force of muscle contraction.

<h3>What is the resultant force due to the quadriceps?</h3>

The resultant of more than two vector forces is given by:

F = √Fₓ² + Fₙ²

where:

Fₓ is the sum of the horizontal components of the forces
Fₙ is the sum of the vertical components of the forces
Fx = F₁cosθ + F₂cosθ + F₃cosθ + F₄cosθ
Fₙ = F₁sinθ + F₂sinθ + F₃sinθ + F₄sinθ
F₁ = 680N, θ = 90 = 30 = 120°
F₂ = 220 N, θ = 90 + 16 = 106°
F₃ = 600 N, θ = 90 + 15 = 105°
F₄ = 480 N, θ = 90 - 35 = 55°

then:

Fx = 680 * cos 120 + 220 * cos 106 + 600 * cos 105 + 480 * cos 55

Fx = -280.6 N

Fₙ = 680 * sin 120 + 220 * sin 106 + 600 * sin 105 + 480 * sin 55

Fₙ = 1773.1 N

then:

F = √(-280.6)² + ( 1773.1)²

F = 1795.16 N

F ≈ 1795 N

Therefore, the resultant force due to the quadriceps is 1795 N

<h3>What would happen if the vastus medialis was trained and strengthened to contract with 720N of force?</h3>

From the new information provided:

F₁ = 680N, θ = 90 = 30 = 120°
F₂ = 220 N, θ = 90 + 16 = 106°
F₃ = 600 N, θ = 90 + 15 = 105°
F₄ = 720 N, θ = 90 - 35 = 55°

then:

Fx = 680 * cos 120 + 220 * cos 106 + 600 * cos 105 + 720 * cos 55

Fx = -142.95 N

Fₙ = 680 * sin 120 + 220 * sin 106 + 600 * sin 105 + 720 * sin 55

Fₙ = 1969.72 N

then:

F = √(-142.95)² + ( 1969.72)²

F = 1974.9 N

F ≈ 1975 N

Therefore, the resultant force due to the quadriceps is 1975 N.

Training and strengthening the vastus medialis results in a greater force of muscle contraction.

Learn more about resultant of forces at: brainly.com/question/25239010

3 0

2 years ago

A parachutist of mass 100 kg falls from a height of 500 m. Under realistic conditions, she experiences air resistance. Based on

antoniya [11.8K]

Given:
mass: 100 kg
height: 500 m
1 kJ = 1000 J
gravity = 9.8 m/s²

velocity before impact: v = √2gh ; v = √2 * 9.8 m/s² * 500 m ; v = 98.99494 m/s

KE = 1/2 m v²
KE = 1/2 * 100 kg * (98.99494 m/s)²
KE = 490,000 J

Pls. see attachment.

5 0

3 years ago

Read 2 more answers