1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Hunter-Best [27]
3 years ago
11

A projectile is launched from ground level at an angle of 30 degrees above the horizontal. Neglect air resistance and consider t

he motion from just after the moment it is launched to just before the moment it lands on the ground. When does the projectiles velocity equal its launch velocity?At the highest point.  The projectile's velocity is never equal to its launch velocity after launch.  The projectile's velocity is always equal to its launch velocity.  Just before landing on the ground.  Halfway back to the ground.  Halfway to the highest point. 
Physics
1 answer:
Oduvanchick [21]3 years ago
6 0

Answer:

just before landing the ground

Explanation:

Let the velocity of projection is u and the angle of projection is 30°.

Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.

initial horizontal component of velocity, ux = u Cos 30

initial vertical component of velocity, uy = u Sin 30

Time of flight is given by

T = \frac{2u Sin\theta }{g}

Final horizontal component of velocity, vx = ux = u Cos 30

Let vy is teh final vertical component of velocity.

Use first equation of motion

vy = uy - gT

v_{y}=u_{y}- g \times \frac{2u Sin\theta }{g}

v_{y}=u Sin 30 - 2u Sin 30

vy = - u Sin 30

The magnitude of final velocity is given by

v = \sqrt{v_{x}^{2}+v_{y}^{2}}

v = \sqrt{\left (uCos 30  \right )^{2}+\left (uSin 30  \right )^{2}}

v = u

Thus, the velocity is same as it just reaches the ground.

You might be interested in
3. How do you think a hot air balloon works?
tiny-mole [99]

Answer:hot air

Explanation:Hot air goes up, so when you put it in a hot air balloon fire is placed at the oppening to creat hot air and lift the balloon! :)

5 0
2 years ago
What time is the eclipse happening tonight?.
andrew11 [14]

The first location to see the partial solar eclipse begin is at 3.58 a.m. EST (08:58 UTC), the greatest point of total solar eclipse occurs at 6 a.m. EST (11:00 UTC) and the last location to see the partial eclipse end is at 8:02 a.m. EST (13:02 UTC) according to Time and Date.

3 0
1 year ago
What is kelvin scale?
Darina [25.2K]
<span>a scale of temperature with absolute zero as zero, and the triple point of water as exactly 273.16 degrees.</span>
4 0
3 years ago
Read 2 more answers
11 A motor is used to lift a load of 40N
RSB [31]

Answer:

The correct answer is option D

4 0
2 years ago
At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
Other questions:
  • What is a characteristic of a default static route? ​?
    10·1 answer
  • A 15.0-kilogram mass is moving at 7.50 meters per second on a horizontal, frictionless surface. What is the
    8·1 answer
  • A positive charge on an object is caused by:
    15·2 answers
  • Choose all correct sentences
    15·1 answer
  •  While texting his girlfriend that he was running late for their date, Jordan rear-ended a car stopped at a red light. Since thi
    6·2 answers
  • Who is this guy and what did he do?
    6·1 answer
  • Light enters an equilateral triangle prism in a direction parallel to one side. The prism is made of glass with an index of refr
    7·1 answer
  • Goal posts at the ends of football fields are padded as a safety measure for players who might run into them. How does thick pad
    15·2 answers
  • A trip is taken that passes through the following points in order
    13·1 answer
  • In what way is a screw similar to an inclined plane?
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!