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
Rudiy27
2 years ago
15

Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?

Physics
1 answer:
frosja888 [35]2 years ago
3 0

Answer:

The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

Explanation:

From the question given above, the following data were obtained:

Height to which the target is located = 50 m

Initial velocity (u) = 20 m/s

To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:

Initial velocity (u) = 20 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 20² – (2 × 10 × h)

0 = 400 – 20h

Collect like terms

0 – 400 = – 20h

– 400 = – 20h

Divide both side by – 20

h = – 400 / – 20

h = 20 m

Thus, the the maximum height to which the cannon ball attained is 20 m.

From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

You might be interested in
If the velocity of a body changes from 13 m/s to 30 m/s while undergoing constant acceleration, what's the average velocity of t
Ronch [10]

Since the acceleration is constant, the average velocity is simply the average of the initial and final velocities of the body:

v_{avg} = \frac{v_f+v_i}{2}=\frac{30 m/s+13 m/s}{2}=21.5 m/s

We can proof that the distance covered by the body moving at constant average velocity v_{avg} is equal to the distance covered by the body moving at constant acceleration a:

- body moving at constant velocity v_{avg}: distance is given by

S=v_{avg}t = \frac{v_f+v_i}{2}t

- body moving at constant acceleration a=\frac{v_f-v_i}{t}: distance is given by

S=v_i t+ \frac{1}{2}at^2 = v_i t + \frac{1}{2}\frac{v_f-v_i}{t}t^2=(v_i+\frac{1}{2}(v_f-v_i))t=\frac{v_f+v_i}{2}t

7 0
2 years ago
Read 2 more answers
Does a satellite have its greatest speed when it is closest to or farthest from earth?
Jet001 [13]
When it is closest to the earth.


3 0
2 years ago
Read 2 more answers
) A 73-mH solenoid inductor is wound on a form that is 0.80 m long and 0.10 m in diameter. A coil having a resistance of is tigh
Aleonysh [2.5K]

Complete question is;. A 73mH solenoid inductor is wound on a form that is 0.80m long and 0.10m in diameter a coil having a resistance of 7.7 ohms is tightly wound around the solenoid at its center the mutual inductance of the coil and solenoid is 19μH at a given instant the current in the solenoid is 820mA and is decreasing at the rate of 2.5A/s at the given instant what is the induced current in the coil

Answer:

6.169 μA

Explanation:

Formula for induced EMF is given by the equation;

EMF = M(di/dt). We are given;

di/dt = 2.5 A/s

M = 19μH = 19 × 10^(-6) H

Thus;

EMF = 19 × 10^(-6) × 2.5.

EMF = 47.5 × 10^(-6) V

Formula for current is;

i = EMF/R. R is resistance given as 7.7 ohms.

Thus; i = 47.5 × 10^(-6)/7.7

i = 6.169 μA

5 0
2 years ago
I need some advice I have two girlfriends as of now.
Paul [167]

Answer:

I Would go with Kye.

Explanation:

Why i would go with her is because she has more life experience with you. She also knows you better. I would usually go with the closest one and the one who knows you the best.

Hope this helps.

<3 Have a good day!

4 0
2 years ago
Read 2 more answers
What are the physical and chemical properties of sodium?
Hitman42 [59]

Answer:

Physical Properties of Sodium

Atomic number 11

Melting point 97.82°C (208.1°F)

Boiling point 881.4°C (1618°F)

Volume increase on melting 2.70%

Latent heat of fusion 27.0 cal/g

Lenntech Water treatment & purification

Toggle navigation

Home Periodic table Elements Sodium

Sodium - Na

Chemical properties of sodium - Health effects of sodium - Environmental effects of sodium

Atomic number

11

Atomic mass

22.98977 g.mol -1

Electronegativity according to Pauling

0.9

Density

0.97 g.cm -3 at 20 °C

Melting point

97.5 °C

Boiling point

883 °C

Vanderwaals radius

0.196 nm

Ionic radius

0.095 (+1) nm

Isotopes

3

Electronic shell

[Ne] 3s1

Energy of first ionisation

495.7 kJ.mol -1

8 0
2 years ago
Other questions:
  • A 500N person stands in an elevator that is moving downward at constant speed. The force that the floor exerts on the person mus
    9·2 answers
  • A baseball goes from zero to 34 m/s in 0.188 s. What is its average acceleration? Answer in units of m/s^2
    15·1 answer
  • Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 25 cm apart. The sound
    9·2 answers
  • In what direction does an applied force move an object?
    5·2 answers
  • The gravitational acceleration on the surface of the moon is 1/6 what it is on earth. If a man could jump straight up 0.7 m (abo
    13·1 answer
  • Obtain the zeroes of polynomial
    8·1 answer
  • A student carries a backpack for one mile , another student carries the same back pack for two miles . Compared to the first stu
    8·2 answers
  • How does the construction of dams positively affect natural resources?
    14·1 answer
  • Which of the following statements are true?
    15·2 answers
  • A science student observes three lines in an emission spectrum, one red, one yellow, and one blue. Which one was caused by an el
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!