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
givi [52]
3 years ago
6

A cannonball is fired across a flat field at an angle of 43 degrees with an initial speed 32 m/s and height of 12 m.

Physics
1 answer:
eimsori [14]3 years ago
3 0

1) x= v_{0x} t = 23.4 t\\y=y_0 + v_{0y}t-\frac{1}{2}gt^2 = 12+21.8t -4.9t^2

The initial data of the projectile are:

y_0 = 12 m is the initial height

v_0 = 32 m/s is the initial speed of the projectile, so its components along the x- and y- directions are

v_{0x} = v_0 cos \theta = (32 m/s)(cos 43^{\circ})=23.4 m/s\\v_{0y} = v_0 sin \theta = (32 m/s)(sin 43^{\circ})=21.8 m/s

The motion of the cannonball along the x-direction is a uniform motion with constant speed, while on the y-direction it is an uniformly accelerated motion with constant acceleration g=9.8 m/s^2 downward. So, the two equations of motion of the projectile along the two directions are:

x= v_{0x} t = 23.4 t\\y=y_0 + v_{0y}t-\frac{1}{2}gt^2 = 12+21.8t -4.9t^2

2) 4.94 s

To determine how long the cannon ball was in the air, we need to find the time t at which the cannonball hits the ground, so the time t at which y(t)=0:

0=12+21.8t-4.9 t^2

Solving the equation with the formula, we have:

t_{1,2}=\frac{-21.8\pm \sqrt{(21.8)^2-4(-4.9)(12)}}{2(-4.9)}

which has two solutions:

t = -0.50 s

t = 4.94 s

Discarding the first solution which is a negative time so it has no physical meaning, the correct solution is

t = 4.94 s

3) 115.6 m

To determine how far the cannonball travelled, we need to find the value of the horizontal position x(t) when the ball hits the ground, at t=4.94 s. Substituting this value into the equation of motion along x, we find:

x=v_{0x}t=(23.4 m/s)(4.94 s)=115.6 m

4) 2.22 s

The cannonball reaches its maximum height when the vertical velocity becaomes zero.

The vertical velocity at time t is given by

v_y(t)= v_{0y} -gt

where

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

Substutiting v_y(t)=0 and solving for t, we find

t=\frac{v_{0y}}{g}=\frac{21.8 m/s}{9.8 m/s^2}=2.22 s

5) 36.2 m

The maximum height reached by the cannon is equal to the vertical postion y(t) when the vertical velocity is zero, so when t=2.22 s. Substituting this value into the equation of the vertical motion, we find:

y(t)=y_0 + v_{0y}t-\frac{1}{2}gt^2=12+(21.8)(2.22)-(4.9)(2.22)^2=36.2 m

You might be interested in
A 5.30 g bullet moving at 963 m/s strikes a 610 g wooden block at rest on a frictionless surface. The bullet emerges, traveling
Tems11 [23]

Answer:

a) V_{wf} = 4.67m/s

b) V = 8.29 m/s

Explanation:

Givens:

The bullet is 5.30g moving at 963m/s and its speed reduced to 426m/s. The wooden block is 610g.

a) From conservation of linear momentum

Pi = Pf

m_{b}V{b_{i} }  + V_{wi}  = m_{w} V_{wf} + m_{b}V_{bf}

where m_{b},V{b_{i} are the mass and the initial velocity of the bullet, m_{w} and V_{wi} are the mass and the initial velocity of the wooden block, and V_{wf} and V_{bf} are the final velocities of the wooden block and the bullet

The wooden block is initial at rest (V_{wi} = 0) this yields

m_{b}V{b_{i} }  = m_{w} V_{wf} + m_{b}V_{bf}

By solving for V_{wf} adn substitute the givens

V_{wf} = \frac{m_{b}V_{bi} - m_{b} V_{bf}    }{m_{W} }

= \frac{5.3(g)(963(m/s)-426(m/s) }{610(g)}

V_{wf} = 4.67m/s

b) The center of mass speed is defined as

V = \frac{m_{b} }{m_{b}+m_{w} } V_{bi}

substituting:

V = \frac{5.3(g)}{5.3(g)+610(g)} X 963(m/s)

V = 8.29 m/s

7 0
3 years ago
The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =
luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time = \frac{\textup{Distance}}{\textup{Velocity}}

or

Time = \frac{\textup{0.016}}{\textup{23}}

or

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

or

a = \frac{q\times8.20\times10^4}{1\times10^{-11}}

Now from the Newton's equation of motion

d=ut+\frac{1}{2}at^2

where,  

d is the distance

u is the initial speed  

a is the acceleration

t is the time

or

0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2

or

q = 9.98 × 10⁻⁹ C

4 0
3 years ago
Which one of the following scenarios accurately describes a condition in which resonance can occur? A. A vibrating tuning fork i
Digiron [165]
I believe the answer is A
5 0
3 years ago
Two point charges each have a value of 3.0 c and are separated by a distance of 4.0 m. what is the electric field at a point mid
swat32
 <span>Place a test charge in the middle. It is 2cm away from each charge. 
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point. 
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out. 
THIS IS A TRICK QUESTION. 
THe electric field exactly midway between them = 0/Q = 0. 
But if the point moves even slightly you need the following formula 
F= (1/4Piε)(Q1Q2/D^2) 
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
4 0
3 years ago
The main difference between a chest and a bounce pass is what?
snow_lady [41]

Answer: The main difference between the three is the mode of transmission. The chest pass is straight through the air towards your teammate. While the bounce pass is directed toward the ground and then at your teammate. Finally, the overhead pass is projected high in the air to avoid defenders.

Explanation:

4 0
2 years ago
Read 2 more answers
Other questions:
  • A table of mass 10 kg is lifted so that the gravitational potential energy of the table increases by 1470 J. How high is the tab
    14·2 answers
  • For a ship moving against the current, it takes 9 hours to cover a distance of 113.4 miles. how much does it take this ship to r
    14·2 answers
  • A transformer connected to a 120-V (rms) ac line is to supply 12,500 V (rms) for a neon sign. To reduce shock hazard, a fuse is
    9·1 answer
  • A wave with high amplitude _____.
    11·2 answers
  • If bert the bat traveles eastward at 40 mph with a tail wind of 6 mph,what is his actual speed?
    5·1 answer
  • Um relógio de ponteiros funciona durante um mês. Qual é, em N.C, o número de voltas, aproximadamente, que o ponteiro dos minutos
    15·1 answer
  • As an object in motion becomes heavier, its kinetic energy _____. A. increases exponentially B. decreases exponentially C. incre
    13·2 answers
  • A double-slit experiment is performed with light of wavelength 620 nm. The bright interference fringes are spaced 2.3 mm apart o
    5·1 answer
  • 46. Can you take a walk in such a way that the distance
    15·1 answer
  • Question 2 A horizontal line on a position vs time graph means the object is O moving faster. O at rest O slowing down. O moving
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!