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Andrej [43]
3 years ago
5

4.5. During a snowball fight, your opponent distracts you by throwing a snowball at you with a high arc. She throws snowballs wi

th a speed of 23.3 m/s and the first is thrown at 70.0 degrees. As you are watching the first snowball, she throws a second at a lower angle. (a) If both snowballs cover the same horizontal distance, at what angle (degrees) should the second be thrown? (b) If the second snow ball is going to hit you while your attention is still focused on the first snowball, how soon (s) after the first snowball is thrown must the second be thrown? In other words, what is the delay between the two throws that will have both arriving at the same time?
Physics
1 answer:
igomit [66]3 years ago
6 0

Answer:

(a) θ = 20°

(b) Time Delay = 2.84 s

Explanation:

(a)

This is the case of the projectile motion. In projectile motion, for the same launch speed, the range of a projectile is the same for the complimentary launch angles. Hence, the snowball will have the same range when launched at 70°, at an angle of:

\theta = 90^{0} - 70^{0}

<u>θ = 20°</u>

<u></u>

(b)

The time of flight of snowball can be found by the following formula:

T = \frac{2\ u\ Sin \theta}{g}

where,

T = Time of flight = ?

u = launch speed = 23.3 m/s

g = acceleration due to gravity = 9.81 m/s²

For θ = 70° :

T_{70} = \frac{(2)(23.3\ m/s)Sin70^{0}}{9.81\ m/s^{2}}

T₇₀ = 4.46 s

For θ = 20° :

T_{20} = \frac{(2)(23.3\ m/s)Sin20^{0}}{9.81\ m/s^{2}}

T₂₀ = 1.62 s

Therefore, the time delay can be calculated as follows:

Time\ Delay = T_{70} - T_{20}\\Time\ Delay = 4.46\ s - 1.62\ s\\

<u>Time Delay = 2.84 s</u>

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Anestetic [448]

Answer:

a. -39.2 m/s; -78.4 m

b. -31.2 m/s; -46.4 m

c. -47.2 m/s; -110.4 m

Explanation:

<h2>Part (a)</h2>

We are given/can infer these variables:

  • t = 4.0 s
  • a = -9.8 m/s²
  • v_0 = 0 m/s

We want to find the displacement and the final velocity of the rock.

  • Δx = ?
  • v = ?

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

  • v = 0 + (-9.8)(4.0)
  • v = -9.8 * 4.0
  • v = -39.2 m/s

The final velocity of the rock is -39.2 m/s.

Now we can use this equation to find the displacement of the rock:

  • Δx = v_0 t + 1/2at²

Plug in the known variables into this equation.

  • Δx = 0 * 4.0 + 1/2(-9.8)(4.0)²
  • Δx = 1/2(-9.8)(4.0)²
  • Δx = -4.9 * 16
  • Δx = -78.4 m

The displacement of the rock is -78.4 m.

<h2>Part (b)</h2>

We are given/can infer these variables:

  • v_0 = 8.0 m/s
  • a = -9.8 m/s²
  • t = 4.0 s

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

  • v = 8.0 + (-9.8)(4.0)
  • v = 8.0 + -39.2
  • v = -31.2 m/s

The final velocity of the rock is -31.2 m/s.

We can use this equation to find the displacement:

  • Δx = v_0 t + 1/2at²

Plug in known variables:

  • Δx = 8.0(4.0) + 1/2(-9.8)(4.0)²
  • Δx = 32 - 4.9(16)
  • Δx = -46.4 m

The displacement of the rock is -46.4 m.

<h2>Part (c)</h2>

We are given/can infer these variables:

  • v_0 = -8.0 m/s
  • a = -9.8 m/s²
  • t = 4.0 s

We can use this equation to find the final velocity:

  • v = v_0 + at

Plug in the known variables into this equation.

  • v = -8.0 + (-9.8)(4.0)
  • v = -8.0 - 39.2
  • v = -47.2 m/s

The final velocity of the rock is -47.2 m/s.

We can use this equation to find the displacement:

  • Δx = v_0 t + 1/2at²

Plug in known variables:

  • Δx = -8.0(4.0) + 1/2(-9.8)(4.0)²
  • Δx = -32 - 4.9(16)
  • Δx = -110.4 m

The displacement of the rock is -110.4 m.

8 0
3 years ago
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