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yawa3891 [41]
3 years ago
12

A ball is thrown down vertically with an initial speed of v0 from a height of h. (a) What is its speed just before it strikes th

e ground? (b) How long does the ball take to reach the ground? What would be the answers to (c) part a and (d) part b if the ball were thrown upward from the same height and with the same initial speed? Before solving any equations, decide whether the answers to (c) and (d) should be greater than, less than, or the same as in (a) and (b).
Physics
1 answer:
Alexus [3.1K]3 years ago
3 0

Answer:

a)   v² = v₀² + 2 g h,  b)   t = v₀/g  (1+ √ (1 + 2gh/ v₀²))

Explanation:

a) This is an exercise that we can solve using conservation of energy.

Starting point. High point

         Em₀ = K + U = ½ m v₀² + m gh

Final point. Soil

         Em_{f} = K = ½ m v²

energy is conserved because there is no friction

         Emo = Em_{f}

         ½ m v₀² + m g h = ½ m v²

         v² = v₀² + 2 g h

b) the time it takes to reach the ground can be calculated with kinematics

let's create a reference frame with positive upward direction

         v = vo - g t

when it reaches the ground it has a velocity v, the initial velocity is downwards v₀ = -v₀

        v = -v₀ - gt

        t = - (v + v₀) / g

we substitute the velocity values ​​calculated in the previous part

        t = - (√(v₀² + 2 g h) + vo) / g

we will simplify the equation a bit

        t = - v₀/g  (1+ √ (1 + 2gh/ v₀²))

c) is now thrown vertically upward with the same initial velocity vo.

   To find the final velocity we use the conservation of energy where the velocity is squared, so it does not matter if it is positive or negative, therefore in this section the value should be the same as in part a

         v = √ (v₀² + 2gh)

d) for this part if there is change since the speed is not squared

     v₀ = v₀

          v = v₀ - gt

          t = (v₀ - v) / g

          t = (v₀ - √(v₀² + 2 g h)) / g

          t = v₀/g   (1 - √(1 + 2gh / v₀²))

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In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st
bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

  • Total time will be the sum of the partial times between stations plus the time stopped at the stations.
  • Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
  • At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
  • In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

       v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s  (1)

  • Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

       t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)

  • Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

       x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m  (3)

  • In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

       t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)

  • We can find the distance traveled while the train was decelerating as follows:

       x_{3} = (v_{1} * t_{3})   + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m  (5)

  • Finally, we need to know the time traveled at constant speed.
  • So, we need to find first the distance traveled at the constant speed of 26.4m/s.
  • This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
  • x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s   (7)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
  • Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
  • tm = t*5 = 131.6 * 5 = 658.2 s (9)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
  • Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

  • Using all the same premises that for a) we know that the only  difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
  • Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
  • We can find the distance traveled at constant speed, rewriting (6) as follows:
  • x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s   (12)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
  • Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
  • tm = t*3 = 207.4 * 3 = 622.2 s (14)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
  • Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)
7 0
3 years ago
Suppose a 4.0-kg projectile is launched vertically with a speed of 8.0 m/s. What is the maximum height the projectile reaches?
eduard

Answer:

h = 3.3 m (Look at the explanation below, please)

Explanation:

This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy

Kinetic energy = \frac{1}{2}mv^{2}

Plug in the numbers = \frac{1}{2}(4.0)(8^{2})

Solve = 2(64) = 128 J

Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.

Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)

Kinetic energy = Potential Energy

128 J = 39.2h

h = 3.26 m

h= 3.3 m (because of significant figures)

7 0
3 years ago
A boy weighs 40 kilograms. He runs at a velocity of 4 meters per second north. Which is his momentum?
Bogdan [553]
The boy's momentum is 160 kg*m/s north.

The formula of momentum is p = mv, where p is momentum.
p = 40 kg * 4m/s north
p =160 kg*m/s north

<span>Thank you for posting your question. I hope you found what you were after. Please feel free to ask me more.</span>


4 0
3 years ago
Read 2 more answers
Using the results of Question 1 that would apply if the collision were inelastic, compute (using Excel in the yellow highlighted
ch4aika [34]

Answer:

The fractional kinetic energy will be lost if the collision is inelastic. In inelastic collision, the kinetic energy is converted into other forms of energy.

The lost energy became heat and sound energy.

Explanation:

During inelastic collision, the kinetic energy of a moving object does not conserve. It changes into another form of energy such as sound energy and heat energy etc.

For example, when a moving car hit another car or wall etc, the kinetic energy is converted into sound and heat energy. This type of collision is inelastic collision.

4 0
3 years ago
1. Simplify the following expression<br> 8-6/4-12+3^2
DanielleElmas [232]

Answer:

-5/2

Explanation:

8-6/4-12+3^2

8-6/4-12+9

2/4-12+9

LCM =4

(2-48+36)/4

(-46+36)/4

-10/4

-5/2

5 0
3 years ago
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