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Gennadij [26K]
2 years ago
10

16. Driverless cars have already , and they look so cool.

Engineering
2 answers:
mash [69]2 years ago
8 0

Answer:C.exist

hope it helps

Gekata [30.6K]2 years ago
4 0
Answer:
C. Exist
Hope it helps!
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Tech A says that water soaked brake shoes can be a cause of brake fade. Tech B says that disc brakes dissipate heat faster than
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Explanation:

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While playing a game of catch on the quadrangle, you throw a ball at an initial velocity of 17.6 m/s (approximately 39.4 mi/hr),
MAXImum [283]

Answer:

a) The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) The ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

Explanation:

a) The ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical motion at constant acceleration. First, we calculate the time taken by the ball to hit the ground:

y = y_{o} + (v_{o}\cdot \sin \theta) \cdot t+\frac{1}{2}\cdot g\cdot t^{2} (1)

Where:

y_{o}, y - Initial and final vertical position, measured in meters.

v_{o} - Initial speed, measured in meters per second.

\theta - Launch angle, measured in sexagesimal degrees.

g - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that y_{o} = 2\,m, y = 0\,m, v_{o} = 17.6\,\frac{m}{s}, \theta = 50^{\circ} and g = -9.807\,\frac{m}{s^{2}}, then the time taken by the ball is:

-4.904\cdot t^{2}+13.482\cdot t +2 = 0 (2)

This second order polynomial can be solved by Quadratic Formula:

t_{1} \approx 2.890\,s and t_{2} \approx -0.141\,s

Only the first root offers a solution that is physically reasonable. That is, t \approx 2.890\,s.

The vertical velocity of the ball is calculated by this expression:

v_{y} = v_{o}\cdot \sin \theta +g\cdot t (3)

Where:

v_{o,y}, v_{y} - Initial and final vertical velocity, measured in meters per second.

If we know that v_{o} = 17.6\,\frac{m}{s}, \theta = 50^{\circ}, g = -9.807\,\frac{m}{s^{2}} and t \approx 2.890\,s, then the final vertical velocity is:

v_{y} = -14.860\,\frac{m}{s}

The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) From a) we understand that ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball (x) is determined by the following expression:

x = (v_{o}\cdot \cos \theta)\cdot t (4)

If we know that v_{o} = 17.6\,\frac{m}{s}, \theta = 50^{\circ} and t \approx 2.890\,s, then the distance covered by the ball is:

x = 32.695\,m

The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before hitting the ground (v), measured in meters per second, is determined by the following Pythagorean identity:

v = \sqrt{(v_{o}\cdot \cos \theta )^{2}+v_{y}^{2}} (5)

If we know that v_{o} = 17.6\,\frac{m}{s}, \theta = 50^{\circ} and v_{y} = -14.860\,\frac{m}{s}, then the magnitude of the velocity of the ball is:

v \approx 18.676\,\frac{m}{s}.

The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is defined by the following trigonometric relationship:

\tan \theta = \frac{v_{y}}{v_{o}\cdot \cos \theta_{o}}

If we know that v_{o} = 17.6\,\frac{m}{s}, \theta_{o} = 50^{\circ} and v_{y} = -14.860\,\frac{m}{s}, the angle of the total velocity of the ball just before hitting the ground is:

\theta \approx -52.717^{\circ}

The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

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