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1 month ago

Can someone please help a struggling physics student?

Physics

1 answer:

Keith_Richards [3.2K]1 month ago

5 0

Part A:

What is the greatest height the ball will achieve?

Utilized kinematics formula:

$v_f^2=v_i^2+2ad$ , where $v_f$ signifies final velocity, $v_i$ denotes initial velocity, $a$ represents acceleration, and $d$ indicates distance traveled. In SI dimensions, velocity should be expressed in $m/s$ , acceleration in $m/s^2$ , and distance in $m$

It is noted that the initial launch velocity is 12.0 m/s upward. Upon reaching the peak height, the vertical speed of the ball will be 0 m/s; otherwise, it will not be apexing. This serves as our final velocity.

The singular acceleration within the context is that due to gravity, which is approximately $9.8\:\mathrm{ m/s^2}$ . The acceleration directed downward, while the ball ascends. For directional clarity, the acceleration must be represented as $-9.8\:\mathrm{m/s^2}$ . We possess three variables and seek to determine the fourth, which is the height. This will indicate the maximum height of the ball.

Substitute $v_i=12$ , $v_f=0$ , $a=-9.8$ to identify $d$ :

$0^2=12^2+2(-9.8)(d),\\0=144-19.6d,\\-19.6d=-144,\\d=\frac{-144}{-19.6}=7.34693877551\approx \boxed{7.35\text{ m}}$

What is the velocity of the ball upon impact with the ground?

This inquiry assesses a concept in physics rather than a specific formula. The vertical velocity upon impacting the ground will equal numerically but be inversely directed compared to the initial vertical velocity. This correlation occurs since the ball invests an equal duration ascending to maximum height as it does returning to the ground (the ball accelerates from initial velocity to 0 and then conversely from 0 to another velocity over the equivalent distance and duration). Given the initial vertical velocity of +12.0 m/s, the speed upon hitting the ground will be $\boxed{-12.0\text{ m/s}}$ . (The negative sign indicates direction. This is essential since velocity is a vector.)

Part B:

**Due to the character limit, I've added Part B's initial question as an image. Please consult the attached image for the response and explanation regarding Part B's first question. I apologize for any inconvenience.**

What is the velocity direction of the ball upon impact? Present your response in terms of the angle (in degrees) of the ball's velocity concerning the horizontal direction (refer to the provided figure).

This question engages a concept similar to that in part A's second question. The ball's launch vertical velocity is numerically equal but inversely directed to its final impact velocity. The horizontal velocity component remains consistent in both magnitude and direction throughout the entire launch since no horizontal forces intervene. Thus, the angle below the horizontal upon impact equates to the launch angle concerning the horizontal direction.

To achieve this, basic trigonometry must be applied for a right triangle setup. In any right-angle triangle, the tangent (tan) of an angle corresponds to the ratio of the opposite side to the adjacent side.

Let the launch angle with the horizontal be $\theta$ . The angle's opposite side corresponds to the vertical launch velocity (12.0 m/s), and the adjacent side represents the horizontal launch velocity (2.3 m/s). Hence, we formulate the following equation:

$\tan \theta=\frac{12.0\text{m/s}}{2.3\text{ m/s}}$

Taking the inverse tangent of both sides:

$\arctan (\tan \theta)=\arctan (\frac{12.0}{2.3})$

Simplifying via $\arctan(\tan \theta)=\theta \text{ for }\theta \in (-90^{\circ}, 90^{\circ})$ :

$\theta=\arctan(\frac{12.3}{2.3}),\\\theta =79.14989537\approx \boxed{79.15^{\circ}}$

We can express our answer, indicating that the velocity direction of the ball upon impacting the ground is $\boxed{\text{approximately }79.15^{\circ} \text{ below the horizontal}}$ or $\boxed{\text{approximately }-79.15^{\circ} \text{ to the horizontal}}$ .

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Incomplete query. The complete inquiry is as follows

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Response:

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Analysis:

Given Information

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Revolutions =3000 rpm

To determine

T( torque )=?

Process

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1 month ago

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