_____ is the kinetic energy of an object, proportional to the object's moment of inertia and the square of its angular velocity.

The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21

frutty [35]

Answer:

a) The velocity of the projectile at 2 seconds after launch is 1.9 meters per second. The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) The projectile reaches maximum height 2.192 seconds after launch.

c) The maximum height of the projectile is 26.584 meters above ground.

d) The projectile will hit the ground at 4.523 seconds after launch.

e) The velocity of the projectile right before hitting the ground in -22.871 meters per second.

Explanation:

Complete statement of problem is: The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21.5 meters per second is $h(t) = 3+21.5\cdot t-4.9\cdot t^{2}$ after t seconds. (Round your answers to two decimal places.) (a) Find the velocity after 2 seconds and after 4 seconds, (b) When does the projectile reach its maximum height? (c) What is the maximum height? (d) When does it hit the ground? (e) With what velocity does it hits the ground?

a) From Physics and Differential Calculus we remember that velocity is the first derivative of height. Hence, we need to differentiate the height function in time:

$v(t) = 21.5-9.8\cdot t$ (Eq. 1)

Where $v(t)$ is the velocity function, measured in meters per second.

Now we evaluate this function at given times:

t = 2 s.

$v(2) = 21.5-9.8\cdot (2)$

$v(2) = 1.9\,\frac{m}{s}$

The velocity of the projectile at 2 seconds after launch is 1.9 meters per second.

t = 4 s.

$v(4) = 21.5-9.8\cdot (4)$

$v(4) = -17.7\,\frac{m}{s}$

The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) Maximum height is reached when velocity of projectile is zero. We equalize velocity to zero and solve the expression for $t$ :

$21.5-9.81\cdot t = 0$

$t = 2.192\,s$

The projectile reaches maximum height 2.192 seconds after launch.

c) Maximum height is calculated by evaluating height function at the time found in b). That is:

$h(2.192) = 3+21.5\cdot (2.192)-4.9\cdot (2.192)^{2}$

$h (2.192) = 26.584\,m$

The maximum height of the projectile is 26.584 meters above ground.

d) In this case, we need to equalize the height function to zero and solve for $t$ . That is:

$3+21.5\cdot t-4.9\cdot t^{2} = 0$

Roots are found by means of Quadratic Formula:

$t_{1}\approx 4.523\,s$ and $t_{2}\approx -0.135\,s$

Only the first root offers a physically reasonable solution. Therefore, the projectile will hit the ground at 4.523 seconds after launch.

e) This can be found by evaluating velocity function at the time found in d):

$v(4.523) = 21.5-9.81\cdot (4.523)$

$v(4.523) = -22.871\,\frac{m}{s}$

The velocity of the projectile right before hitting the ground in -22.871 meters per second.

5 0

4 years ago

True or False:In a current, electrons will always flow from negative to positive.

DIA [1.3K]

It should be true: electrons low from negative toward positive to negative toward positive because opposite charges attract each other.

I hope this was correct

3 0

3 years ago

Read 2 more answers

You name binary ionic compounds by naming the first element or the ___ first,

Nuetrik [128]

Answer:

cation, anion

Explanation:

Binary Ionic Compounds- Binary means 2, ionic relates to charges. Therefore binary ionic compounds have TWO elements, with charges that cancel eachother out. The cation (or positive charge) goes before the anion (or negative charge)

example:

NaCl

Na^+1

Cl^-1

Sodium Chloride

6 0

4 years ago

Which of the following determines the range of spectral lines produced during electron transition?. . A.The total number of ener

KengaRu [80]

Answer: A.The total number of energy levels the electron can jump to.

Explanation:

Spectral lines are bright or dark lines over continuous spectrum which occur due to emission or absorption of energy.

When an electron jumps to or from one energy level to another energy level, spectral lines are produced. The range of spectral lines depends on the number of energy levels available to which the electron can jump. This depends the amount of energy gained/lost by the electron.

Thus, the correct answer is: A.The total number of energy levels the electron can jump to.

8 0

3 years ago

The Indianapolis speedway consists of a 2.5 mile track having four turns, each 0.25 mile long and banked at 9 12'

blsea [12.9K]

Answer: Your question is missing below is the question

Question : What is the no-friction needed speed (in m/s ) for these turns?

answer:

20.1 m/s

Explanation:

2.5 mile track

number of turns = 4

length of each turn = 0.25 mile

banked at 9 12'

Determine the no-friction needed speed

First step : calculate the value of R

2πR / 4 = πR / 2

note : πR / 2 = 0.25 mile

∴ R = ( 0.25 * 2 ) / π

= 0.159 mile ≈ 256 m

Finally no-friction needed speed

tan θ = v^2 / gR

∴ v^2 = gR * tan θ

v = √9.81 * 256 * tan(9.2°) = 20.1 m/s

8 0

3 years ago