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2) an astronaut drops a feather from 1.2 meters above the surface of th…

Question

  1. an astronaut drops a feather from 1.2 meters above the surface of the moon. if the acceleration of gravity on the moon is 1/6 the acceleration on earth, how long does it take the feather to hit the moon?
  2. a stone falls freely from rest for 8 seconds.

a) find vf.
b) find d

  1. during a baseball game, a batter hits a pop fly. if the ball remains in the air for 6 seconds, how high does it rise?
  2. a student drops a rock from a bridge to the water 12 meters below. what is the speed of the rock when it strikes the water?
  3. an object with an initial velocity of 20.0 cm/s is accelerated at 8.0 cm/s² for 5.0 s.

a) what is the total displacement?
b) what is the displacement during the fifth second?
what velocity is attained by an object that is accelerated at 0.30 m/s² for a distance of 54 m if its initial velocity is 0.50 m/s?

Explanation:

Response

Let's solve each problem one by one using kinematic equations.

Problem 2: Feather on the Moon

We use the kinematic equation for free fall: \( d = v_0 t + \frac{1}{2} a t^2 \). The initial velocity \( v_0 = 0 \) (dropped), \( d = 1.2 \, \text{m} \), and \( a = \frac{1}{6} g \) where \( g = 9.8 \, \text{m/s}^2 \) (earth's gravity). So \( a = \frac{9.8}{6} \approx 1.633 \, \text{m/s}^2 \).

Step1: Identify knowns and equation

\( d = 1.2 \, \text{m} \), \( v_0 = 0 \), \( a = \frac{9.8}{6} \, \text{m/s}^2 \), use \( d = \frac{1}{2} a t^2 \) (since \( v_0 = 0 \)).

Step2: Solve for \( t \)

Rearrange the equation: \( t = \sqrt{\frac{2d}{a}} \)

Substitute \( d = 1.2 \) and \( a = \frac{9.8}{6} \):

\( t = \sqrt{\frac{2 \times 1.2}{\frac{9.8}{6}}} = \sqrt{\frac{2.4 \times 6}{9.8}} = \sqrt{\frac{14.4}{9.8}} \approx \sqrt{1.469} \approx 1.21 \, \text{seconds} \)

Problem 3: Stone falling freely

We use \( v_f = v_0 + a t \) (for part a) and \( d = v_0 t + \frac{1}{2} a t^2 \) (for part b), with \( v_0 = 0 \), \( a = g = 9.8 \, \text{m/s}^2 \), \( t = 8 \, \text{s} \).

Part a: Find \( v_f \)

Step1: Use \( v_f = v_0 + g t \)

\( v_0 = 0 \), so \( v_f = 0 + 9.8 \times 8 = 78.4 \, \text{m/s} \)

Part b: Find \( d \)

Step1: Use \( d = \frac{1}{2} g t^2 \)

\( d = \frac{1}{2} \times 9.8 \times 8^2 = 4.9 \times 64 = 313.6 \, \text{meters} \)

Problem 4: Pop fly in baseball

The time to rise is half the total time in air (since time up = time down). So \( t = \frac{6}{2} = 3 \, \text{s} \). At the peak, \( v_f = 0 \). Use \( v_f = v_0 - g t \) (upward motion, so acceleration is -g) to find initial upward velocity, then use \( v_f^2 = v_0^2 - 2 g d \) to find height \( d \).

Step1: Find \( v_0 \) at launch

At peak, \( v_f = 0 \), \( t = 3 \, \text{s} \), \( a = -g \). So \( 0 = v_0 - 9.8 \times 3 \implies v_0 = 29.4 \, \text{m/s} \)

Step2: Find height \( d \)

Use \( v_f^2 = v_0^2 - 2 g d \). At peak, \( v_f = 0 \), so \( 0 = (29.4)^2 - 2 \times 9.8 \times d \)

\( 2 \times 9.8 \times d = 29.4^2 \implies d = \frac{29.4^2}{2 \times 9.8} = \frac{864.36}{19.6} = 44.1 \, \text{meters} \)

Problem 5: Rock dropped from bridge

Use \( v_f^2 = v_0^2 + 2 a d \), where \( v_0 = 0 \), \( a = g = 9.8 \, \text{m/s}^2 \), \( d = 12 \, \text{m} \).

Step1: Identify equation

\( v_f^2 = 0 + 2 \times 9.8 \times 12 \)

Step2: Solve for \( v_f \)

\( v_f^2 = 235.2 \implies v_f = \sqrt{235.2} \approx 15.33 \, \text{m/s} \)

Problem 6: Object with initial velocity

We use \( d = v_0 t + \frac{1}{2} a t^2 \) (for part a) and displacement during the fifth second is displacement at \( t = 5 \) minus displacement at \( t = 4 \) (for part b). \( v_0 = 20.0 \, \text{cm/s} \), \( a = 8.0 \, \text{cm/s}^2 \), \( t = 5.0 \, \text{s} \).

Part a: Total displacement

Step1: Use \( d = v_0 t + \frac{1}{2} a t^2 \)

\( d = 20.0 \times 5.0 + \frac{1}{2} \times 8.0 \times (5.0)^2 \)

\( d = 100 + 4.0 \times 25 = 100 + 100 = 200 \, \text{cm} \)

Part b: Displacement during fifth second

Step1: Displacement at \( t = 5 \): \( d_5 = 20.0 \times 5 + \frac{1}{2} \times 8.0 \times 25 = 100 + 100 = 200 \, \text{cm} \) (wait, no, earlier calculation was wrong? Wait, \( \frac{1}{2} \times 8 \times 25 = 100 \), so \( 20 \times 5 = 100 \), total 200. Wait, displacement at \( t = 4 \): \( d_4 = 20.0 \times 4 + \frac{1}{2} \times 8.0 \times 16 = 80 + 64 = 144 \, \text{cm} \)

Step2: Displacement in fifth second: \( d_5 - d_4 = 200 - 144 = 56 \, \text{cm} \)

Last Problem: Velocity attained

Use \( v_f^2 = v_0^2 + 2 a d \), where \( v…

Answer:

s:

  1. \( \approx 1.21 \, \text{s} \)
  2. a) \( 78.4 \, \text{m/s} \); b) \( 313.6 \, \text{m} \)
  3. \( 44.1 \, \text{m} \)
  4. \( \approx 15.33 \, \text{m/s} \)
  5. a) \( 200 \, \text{cm} \); b) \( 56 \, \text{cm} \)

Last Problem: \( \approx 5.71 \, \text{m/s} \)