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yawa3891 [41]
3 years ago
13

In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. controlling the angle of the sp

ike is difficult. suppose a ball is spiked from a height of 2.10 m with an initial speed of 17.0 m/s at a downward angle of 15.0°. how much farther on the opposite floor would it have landed if the downward angle were, instead, 7.0°?
Physics
1 answer:
8090 [49]3 years ago
6 0
V ( initial ) = 20 m/s
h = 2.30 m
h = v y * t + g t ² / 2
d = v x * t
1 ) At α = 18°:
v y = 20 * sin 18° = 6.18 m/s
v x = 20 * cos 18° = 19.02 m/ s
2.30 = 6.18 t + 4.9 t²
4.9 t² + 6.18 t - 2.30 = 0
After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ):
t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 )  
t = 0.3 s
d 1 = 19.02 m/s * 0.3 s = 5.706 m
2 ) At  α = 8°:
v y = 20* sin 8° = 2.78 m/s
v x = 20* cos 8° = 19.81 m/s
2.3 = 2.78 t + 4.9 t² 
4.9 t² + 2.78 t - 2.3 = 0
t = 0.46 s
d 2 = 19.81 * 0.46 = 9.113 m
The distance is:
d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m

GOOD LUCK AND HOPE IT HELPS U
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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
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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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