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3 years ago

What angle would you use for the greatest distance when the projectile leaves at zero height above ground g

Physics

2 answers:

fomenos3 years ago

7 0

Answer: 45 degree

Explanation:

Since Range = Ucosø × T

Where T = total time.

Range = distance covered

T = 2usinø/ 2g

If you substitute T into the range formula, you will get

R = (Ucosø×2Usinø) / 2g

But in trigonometry, 2sinøcosø = sin2ø

Substitute it into the formula

R = Usin2ø/ 2g

If ø = 45 degree

R = Usin(2 × 45)/2g

R = Usin90 / 2g

But sin 90 = 1

Therefore,

Range R = U/2g

Therefore, 45 degree angle would be use for the greatest distance when the projectile leaves at zero height above ground.

insens350 [35]3 years ago

5 0

Answer:

∴ the angle for greatest distance is 90°, where Vy = 0

Explanation:

maximum height of a projected object is highest vertical position of the trajectory object which depends on the initial velocity

formula for maximum height

H= u² sin²θ/2g

if θ = 90°, then sin²90° = 1

∴ the angle for greatest distance is 90°, where Vy = 0

for horizontal distance,

the range of a projectile is the horizontal distance.

R = u² sin2θ/g

if θ = 45°, then 2θ = 90°

∴sin 90°= 1

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aivan3 [116]

Answer:

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3 years ago

Suzie Spacewalker hovers in space beside a rotating space station in outer space. Both she and the center of mass of the space s

Inga [223]

Answer:

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Explanation:

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3 years ago

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 turn

kumpel [21]

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:

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.

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

(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

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: