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Elden [556K]
2 years ago
14

PLEASE HELP A student climbs to the top of a press box where the cameras are. He wonders how many meters he is off of the ground

below. So he drops a golfball he had in his pocket, and his friend times how long it takes for the ball to hit the concrete behind the press box. If it takes 1.5 seconds for the ball to hit. What is the height of the rail on top of the press box where the ball was dropped from?
Physics
1 answer:
Feliz [49]2 years ago
3 0

The height of the rail on top of the press box where the ball was dropped from is 11.025 m.

The given parameters:

  • Time of motion of the ball, t = 1.5 s
  • Let the height of the rail = h

<h3>Maximum height of fall;</h3>
  • The maximum height through which the ball was dropped from is calculated by applying second equation of motion;

h = v_0_y t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\h = \frac{1}{2} (9.8) (1.5)^2\\\\h = 11.025 \ m

Thus, the height of the rail on top of the press box where the ball was dropped from is 11.025 m.

Learn more about height of projectiles here: brainly.com/question/10008919

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Answer:

(A) 570 rad

(B) 10 s

(C) 12.5 rad/s²

Explanation:

The equations of motion for circular motions are used.

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  • Angular acceleration, \alpha =35.0 \text{ rad/s}^2

(A)

At <em>t</em> = 2.00 s, the angular displacement, <em>θ</em>, is given by

\theta = \omega_0t+\frac{1}{2}\alpha t^2 = (30\times 2) + \frac{1}{2}\times35\times2^2=60+70 = 130\text{ rad}

After this time, it decelerates through an angular displacement of 440 rad.

Total angular displacement = 130 + 440 rad = 570 rad

(B)

At the time the circuit breaker tips, the angular velocity is given by

\omega = \omega_0+\alpha  t = 30.0+(35.0\times 2) = 30.0+70.0 =100.0\ \text{rad/s}

This becomes the initial angular velocity for the decelerating motion. Because it stops, the final angular velocity is 0 rad/s. The time for this part of the motion is calculated thus:

\theta_2 = \left(\dfrac{\omega_i+\omega_f}{2}\right)t

Here, \theta_2=440 (the angular displacement during deceleration)

The subscripts, <em>i</em> and <em>f</em>, on <em>ω</em> denote the initial and final angular velocities during deceleration.

\omega_i = 100

\omega_f = 0

t = \dfrac{2\theta_2}{\omega_i} = \dfrac{2\times400}{100} = 8\ \text{s}

This is the time for deceleration. The deceleration began at <em>t</em> = 2 s.

Hence, the wheel stops at <em>t</em> = 2 + 8 = 10 s.

(C)

The deceleration is given by

\alpha_R = \dfrac{\omega_f-\omega_i}{t} = \dfrac{0-100}{8} = -12.5\text{ rad/s}^2

The negative sign appears because it is a deceleration.

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Answer:

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Answer:

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