In which of the following wave types is the particle motion perpendicular to wave motion?

A ball is thrown from a rooftop with an initial downward velocity of magnitude vo = 2.9 m/s. The rooftop is a distance above the

Step2247 [10]

Answer:

a) The velocity of the ball when it hits the ground is -20.5 m/s.

b) To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

Explanation:

I´ve found the complete question on the web:

A ball is thrown from a rooftop with an initial downward velocity of magnitude v0=2.9 m/s. The rooftop is a distance above the ground, h= 21 m. In this problem use a coordinate system in which upwards is positive.

(a) Find the vertical component of the velocity with which the ball hits the ground.

(b) If we wanted the ball's final speed to be exactly 27, 3 m/s from what height, h (in meters), would we need to throw it with the same initial velocity?

The equation of the height and velocity of the ball at any time "t" are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the ball at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the ball at a time "t".

First, let´s find the time it takes the ball to reach the ground (the time at which h = 0)

h = h0 + v0 · t + 1/2 · g · t²

0 = 21 m - 2.9 m/s · t - 1/2 · 9.8 m/s² · t²

Solving the quadratic equation using the quadratic formula:

t = 1.8 s ( the other solution of the quadratic equation is rejected because it is negative).

Now, using the equation of velocity, let´s find the velocity of the ball at

t = 1.8 s:

v = v0 + g · t

v = -2.9 m/s - 9.8 m/s² · 1.8 s

v = -20.5 m/s

The velocity of the ball when it hits the ground is -20.5 m/s.

b) Now we have the final velocity and have to find the initial height. Using the equation of velocity we can obtain the time it takes the ball to acquire that velocity:

v = v0 + g · t

-27.3 m/s = -2.9 m/s - 9.8 m/s² · t

(-27.3 m/s + 2.9 m/s) / (-9.8 m/s²) = t

t = 2.5 s

The ball has to reach the ground in 2.5 s to acquire a velocity of 27.3 m/s.

Using the equation of height, we can obtain the initial height:

h = h0 + v0 · t + 1/2 · g · t²

0 = h0 -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

-h0 = -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²

h0 = 38 m

To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.

6 0

4 years ago

A projectile is fired vertically from Earth's surface with an initial speed v0. Neglecting air drag, how far above the surface o

lidiya [134]

This may helpv^2=u^2+2as. v=0 at top of flight. a=acceleration of gravity(vo^2)/2a=s.

3 0

3 years ago

A. If we increase the wind velocity, the maximum vertical dispersal height and rate of diffusion will decrease____.

vovikov84 [41]

Answer:

a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase

b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.

c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase

Explanation:

a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase

b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.

c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase

8 0

3 years ago

Read 2 more answers

PLEASE HELP 100 POINTS! Please fill in the scale distance from sun and diversion factor, listing off the numbers works

steposvetlana [31]

Solved your another question same like this with scaling to Cm this time we go with metre(m)

Scale factor

1km=10³m
1m=10^{-3}km

Mercury

58000m

Ven us

108000m

Earth

150000m

Mars

228000m

Jupiter

778000m

Saturn

1430000m

Uranus

2870000m

Neptune

4500000m

7 0

2 years ago

a truck was traveling at 16.6 miles per second and accelerates at a rate of 2.0 meters per second squared how much time is requi

Ivan

A truck was traveling at 16.6 miles per second and accelerates at a rate of 2.0 meters per second squared then time is required for the truck to reach a speed of 25 miles per second is 6759 s.

Explanation:

Velocity is defined as the rate of change in displacement while acceleration is defined as the rate of change of velocity. Acceleration may be positive or negative. Acceleration is positive when the velocity of the object is increases and it is negative when velocity of the object is decreases. Negative acceleration is also called deceleration.

Mathematically

a = $\frac{(v_{f} - v_{i})}{t}$

Where a is the acceleration of the object, $v_{f}$ is the final velocity of the object and $v_{i}$ is the initial velocity of the object. t is equal to time taken.

Given data:

$v_{f}$ = 25 miles/s

$v_{i}$ = 16.6 miles/s

a = 2.0 m/s²

t = ?

As velocities and acceleration given in different units, So we need to convert to obtain same units. Here we convert unit of acceleration from m/s² to miles/s².

1 m/s² = 0.000621371192 miles/s²

2 m/s² = 0.00124274238 miles/s²

So,

a = 0.00124274238 miles/s²

Apply formula

a = $\frac{v_{f} - v_{i}}{t}$