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3 years ago

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Assume that a projectile is fired horizontally from a "gun" located 1 meter above the ground. If the ball strikes the ground a d

istance of 2 meters from the end of the "gun" determine the muzzle velocity of the gun.

1 answer:

IrinaVladis [17]3 years ago

4 0

Answer:

$v_{o}=8.85m/s$

Explanation:

To determine the <u>muzzle velocity of the gun</u>, we must know how long does it take the ball to <u><em>strikes the ground </em></u>

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

Since the ground is at y=0 and $v_{oy}=0$

$0=1-\frac{1}{2}(9.8)t^{2}$

Solving for t

$t=0.4517s$

Now, to determine the muzzle velocity we need to find its acceleration first

$x=x_{o}+v_{ox}t+\frac{1}{2}at^{2}$ (1)

$v=v_{ox}+at$ (2)

If we analyze the final velocity is 0. From (2) we have that

$v_{ox}=-at$ (3)

Replacing (3) in (1)

$2=-at^{2}+\frac{1}{2}at^{2}$

$2=a(0.4517)^{2} (\frac{1}{2}-1)$

$a=-19.60m/s^{2}$

Solving (3)

$v_{ox}=-at=-(19.60m/s^{2} )(0.4517s)=8.85m/s$

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kotegsom [21]

Answer:

Explanation:

v= u + at

v is final velocity , u is initial velocity . a is acceleration and t is time

Initial velocity u = 0 . Putting the given values in the equation

v = 0 + g sin 18 x 3.5

= 10.6 m /s

3 0

3 years ago

How much momentum does a 10 kg object have when is moving 30 m/s

nikklg [1K]

P = m x v

P = 30 x 10
=300

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4 years ago

A golf ball (m = 59.2 g) is struck a blow that makes an angle of 50.5 ◦ with the horizontal. The drive lands 130 m away on a fla

fomenos

Answer:

The force is calculated as 338.66 N

Explanation:

We know that force is given by

$\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=\frac{m(v_{f}-v_{o})}{\Delta t}$

We know that range of a projectile is given by

$R=\frac{v_{o}^{2}sin(2\theta )}{g}$

it is given that R=130 m applying values in the above equation we get

$R=\frac{v_{o}^{2}sin(2\theta )}{g}\\\\v_{0}=\sqrt{\frac{Rg}{sin(2\theta )}}\\\\\therefore v_{o}=\sqrt{\frac{130\times 9.81}{sin(2\times 50.5_{o}} )}\\\\v_{o}=36.04m/s$

Thus the force is obtained as

$\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=\frac{59.2\times 10^{}-3(36.04-0)}{6.3\times 10^{-3}}$

Thus force equals $F=338.66N$

4 0

4 years ago

Read 2 more answers

The auto in the sketch moves forward as the brakes are applied. A bystander says that during the interval of braking, the auto's

Ivan

Answer:

The statement is true: velocity and acceleration have opposite directions in the interval of braking.

Explanation:

Let's say we have a velocity $v>0$ .

The acceleration $a$ is the rate of change of the velocity $v$ . This means that if $v$ is <em>increasing during</em> time, then $a$ must be positive. But if $v$ is <em>decreasing over</em> time, then $a$ will be negative (even though the velocity is positive).

Mathematically:

$a=\frac{dv}{dt}$

$v$ decreases ⇒ $\frac{dv}{dt}$

⇒ $a$ .

Example:

$v(t)=e^{-t}>0 \\\\\frac{dv}{dt}=-te^{-t}$

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4 years ago

In a hydraulic garage lift, the small piston has a radius of 5.0 cm and the large piston has a radius of 1.50 m. What force must

Nikitich [7]

Answer:

222.22 N

Explanation:

r = 5 cm = 0.05 m, R = 1.5 m, f = ? , F = 200,000 N

Use pascal's law

F / A = f / a

F / 3.14 x R^2 = f / 3.14 x r^2

F / R^2 = f / r^2

f = F x r^2 / R^2

f = 200000 x 0.05 x 0.05 / (1.5 x 1.5)

f = 222.22 N

7 0

3 years ago