Please help!

A projectile is launched horizontally from the top a 35.2m high cliff and lands a distance of 107.6m from the base of the cliff.

tankabanditka [31]

Answer:

$v_o=40.14\ m/s$

Explanation:

<u>Horizontal Launch </u>

It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:

The horizontal speed is constant and equal to the initial speed $v_o$
The vertical speed is zero at launch time, but increases as the object starts to fall
The height of the object gradually decreases until it hits the ground
The horizontal distance where the object lands is called the range

We have the following formulas

$\displaystyle v_x=v_o$

$\displaystyle x=v_o.t$

$\displaystyle v_y=g.t$

$\displaystyle y=\frac{gt^2}{2}$

Where $v_o$ is the initial horizontal speed, $v_y$ is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.

If we know the initial height of the object, we can compute the time it takes to hit the ground by using

$\displaystyle y=\frac{gt^2}{2}$

Rearranging and solving for t

$\displaystyle 2y=gt^2$

$\displaystyle t^2=\frac{2\ y}{g}$

$\displaystyle t=\sqrt{\frac{2\ y}{g}}$

We then replace this value in

$\displaystyle x=v_o.t$

To get

$\displaystyle v_o=\frac{x}{t}$

$\displaystyle v_o=\frac{x}{\sqrt{\frac{2y}{g}}}$

$\displaystyle v_o=\sqrt{\frac{g}{2y}}.x$

The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing $v_o$

$\displaystyle v_o=\sqrt{\frac{9.8}{2(35.2)}}\ 107.6$

The launch velocity is

$\boxed{v_o=40.14\ m/s}$

7 0

3 years ago

A boy and a girl are balanced on a massless seesaw. The boy has a mass of 60 kg and the girl's mass is 50 kg. If the boy sits 1.

skelet666 [1.2K]

Answer: 1.8m

Explanation:

Using the principle of moment to solve the question which states that the sum of clockwise moment is equal to the sum of anticlockwise moments.

The boy moves in the clockwise direction towards the pivot while the girl moves in the anticlockwise direction towards the pivot.

Since moment = force × perpendicular distance

CW moments = 60×1.5 = 90Nm

ACW moments = 50×x = 50x

Equating both moments to get 'x'

90 = 50x

x =9/5

x = 1.8m

This shows that the girl must sit 1.8m from the other side for equilibrium.:

3 0

3 years ago

What is a car's acceleration if it increases its speed from 5 m/s to 20 m/s in 3 s?

liberstina [14]

the correct answer would be C. 5m/s^2

20-5 = 15

15/3 = 5

8 0

3 years ago

Read 2 more answers

An automobile has a vertical radio antenna 1.45 m long. The automobile travels at 70.0 km/h on a horizontal road where Earth's m

Agata [3.3K]

Answer:

595.73 X 10⁻⁶ V.

Explanation:

The magnetic field is towards the north with dip angle equal to 65 so the horizontal component of earth's magnetic field will be

H = 50 X 10⁻⁶ Cos 65 = 21.13 x 10⁻⁶ T.

length of rod L = 1.45 M

Velocity of rod = 70 km / h = 19.444 m /s

Since the top of the antenna rod is to have positive polarity , current induced in it will be from down to up . Magnetic field is towards the north .

To cut magnetic flux perpendicularly , rod should move towards the east .

We can easily get this direction by applying Fleming's right hand Rule.

Induced emf = H L v

= 21.13 x 10⁻⁶ X 1.45 X 19.444

= 595.73 X 10⁻⁶ V.

8 0

3 years ago

An electric pencil sharpener is powered by 6 volts. The current through the pencil sharpener is 0.5 amps. What's is the resistan

Viefleur [7K]

Answer:

Resistance in pencil sharpener = 12 ohms

Explanation:

Given:

Voltage in pencil sharpener = 6 volts

Current in pencil sharpener = 0.5 amps

Find:

Resistance in pencil sharpener

Computation:

Resistance = Voltage / Current

Resistance in pencil sharpener = Voltage in pencil sharpener / Current in pencil sharpener

Resistance in pencil sharpener = 6 / 0.5

Resistance in pencil sharpener = 12 ohms

7 0

3 years ago