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3 years ago

5

A shell if fired from the ground with an initial velocity of 1,700 m/s at an initial angle of 55 degrees to the horizontal. Negl

ecting air resistance, what is the shell's horizontal range?

1 answer:

Triss [41]3 years ago

8 0

Answer:

Therefore the horizontal range = 294897.96 m.

Explanation:

Range of a projectile: The range is defined as the horizontal distance from the point of projection to the point where the projectile hit the projection plane again. The S.I unit of range is Meter (m).

It can be expressed mathematically as

R = u²sin2∅/g............................. Equation 1

Where R = Horizontal range, ∅ = angle of projection, u = initial velocity, g = acceleration due to gravity.

<em>Given: u = 1700 m/s, </em>∅ = 55°,

Constant: g = 9.8 m/s²

Substituting these values into equation 1

R = (1700²sin55)/9.8

R = 2890000/9.8

R = 294897.96 m.

Therefore the horizontal range = 294897.96 m.

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Answer:

a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

Explanation:

Statement is incomplete. The complete description is now described below:

<em>A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off. </em>

<em>(a) What is the magnitude of the satellite's velocity when the thruster turns off</em>

<em>(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis</em>

Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite ( $\vec{v}_{S}$ ), measured in meters per second, is:

$\vec{v}_{S} = (v_{o,x}+a_{x}\cdot t)\,\hat{i}+v_{y}\,\hat{j}$

Where:

$v_{o,x}$ - Initial velocity in +x direction, measured in meters per second.

$a_{x}$ - Acceleration in +x direction, measured in meter per square second.

$t$ - Time, measured in seconds.

$v_{y}$ - Velocity in +y direction, measured in meters per second.

If we know that $v_{o,x} = 0\,\frac{m}{s}$ , $a_{x} = 0.250\,\frac{m}{s^{2}}$ , $t = 45\,s$ and $v_{y} = 21.4\,\frac{m}{s}$ , the final velocity of the satellite is:

$\vec{v}_{S} = \left[0\,\frac{m}{s}+\left(0.250\,\frac{m}{s^{2}} \right)\cdot (45\,s) \right]\,\hat{i}+\left(21.4\,\frac{m}{s} \right)\,\hat{j}$

$\vec{v_{S}} = 11.25\,\hat{i}+21.4\,\hat{j}\,\,\left[\frac{m}{s} \right]$

a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:

$\|\vec {v}_{S}\| = \sqrt{\left(11.25\,\frac{m}{s} \right)^{2}+\left(21.4\,\frac{m}{s} \right)^{2}}$

$\|\vec{v}_{S}\| \approx 24.177\,\frac{m}{s}$

The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:

$\tan \alpha = \frac{v_{y}}{v_{x}} = \frac{21.4\,\frac{m}{s} }{11.25\,\frac{m}{s} }$

$\tan \alpha = 1.902$

$\alpha = \tan^{-1}1.902$

$\alpha \approx 62.266^{\circ}$

The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

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Answer:

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