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1. What is the acceleration of a spaceship that can accelerate from 10,000 m/s to 17,500 m/s in 20 seconds?

2. How long does it take the principal to run to our classroom, if the distance is 125 meters, and he accelerates, from rest, at a rate of 4 m/s^2?

3. A pen falls off my desk. It took 1.5 seconds to fall. What was the velocity of the pen just before it hit the ground?

4. You are standing at the edge of a cliff that is 40m tall. You throw an apple down with a speed of 15 m/s. How long does it take to fall to the bottom of the cliff?

5. A stone is dropped from the roof of a building. It took 5 s for the stone to reach the ground. What is the height of the building?

6. A man is on a 30m cliff and throws a ball down with a speed of 7 m/s.
A) With what velocity does the ball hit the ground? B) How high above the ground is the ball after the ball has fallen for 1 second?

7. A watermelon is dropped from rest from the top of a 20m cliff. A) How long does the watermelon be in the air?
B) What was the watermelon’s final velocity?

The kinematic relations allow finding the results for the different questions are:

     1.  The acceleraations is a = 375 m/s²

     2. The time is t = 62.5 s

     3. The velocity at the ground is  v = -14.7 m / s

     4.  The time to the bottoncliff is    1.7 s

     5.  The height isy₀ = 122.5 m

     6.

         A) The velocity is v = 25.2 m / s

         B)  The height  for t=1 s is  y = 18.1 m

     7.

         A)  Time in the air is t = 2 s

         B)  The final veloicty is   v = - 19.6 m / s

Kinematics analyzes the movement of bodies, finding relationships between the position, velocity and acceleration of bodies.

Let's look for the answers to a series of questions:

1. They indicate the initial and final velocities of the body and the time to reach it is 20 s, ask how much the acceleration is worth

Let's use the relationship

             v = v₀ + a t

             a = [tex]\frac{v-v_o}{t}[/tex]

             a = [tex]\frac{17500-10000}{20}[/tex]

             a = 375 m/s²

2. What is the time it takes to travel a distance of 125 m with an acceleration of 4 m / s²

             x = v₀ t + ½ a t²

As it leaves the direction its initial velocity is zero

             x = ½ to t²

             t = [tex]\sqrt{\frac{2x}{a} }[/tex]

             t = [tex]\sqrt{\frac{2 \ 125}{4} }[/tex]

             t = 62.5 s

3. A pencil falls from a height and reaches the floor in a time of 1.5 s what is the speed when reaching the floor

            v = v₀ - g t

As it is released its initial velocity is zero

            v = -g t

            v = - 9.8 1.5

            v = -14.7 m / s

The negative sign indicates that the velocity is directed downwards.

4. From the edge of a cliff 40 m high, an apple is thrown downward with an initial velocity of 15 m / s. How long does it take to get to the bottom?

             y = y₀ + v₀ t - ½ g t²

In this case the initial velocity is negative because it is directed downwards and when it reaches the floor its height is zero, let us substitute

Suppose the acceleration is g = 10 m / s²

             0 = 40 - 15 t - ½ 10 t²

             0 = 40 - 15 t - 5 t²

let's solve the quadratic equation

             t² + 3 t - 8 = 0

             t = -3 + Ts 9 + 4 8/2

            t = -3 + 6. 4/2

             t₁ = - 4.9 s

            t₂ = 1.7 s

The time must be a positive quantity, so the correct answer is 1.7 s

5. A stone is dropped from the ceiling and it takes a time 5 s to reach the floor, which is the height of the ceiling

           y = y₀ +v₀ t - ½ g t²

as it is released from the ceiling its initial velocity is zero and the height upon reaching the floor is zero

          0 = y₀ + 0 - ½ g t²

          y₀ = ½ g t²

          y₀ = ½ 9.8 5²

          y₀ = 122.5 m

6. A man at a height of 30 m throws a ball downward at 7 m / s

    A) The speed when reaching the ground

    B) The height of the ball when a time of 1 s has fallen.

  A) Let's use the kinematics relation

              v² = v₀² - 2 g (y -y₀)

              v = 7² - 2 9.8 (0 -30)

              v = 25.2 m / s

 B)   y = y₀ + v₀ t - ½ g t²

        y = 30  7 1 - ½ 9.8 1²

        y = 18.1 m

7.  A watermelon is dropped from rest from the top of a 20m cliff.  

 A) time in the air

 B) ground speed

  A) Let's use the relation

       y = y₀ + v₀ t - ½ g t²

It is released its initial velocity is zero and when it reaches the ground its height is zero (y = 0)

       0 = y₀ + 0 - ½ g t²

       t = [tex]\sqrt{\frac{2y_o}{g} }[/tex]

       t = [tex]\sqrt{\frac{2 \ 20}{9.8} }[/tex]

       t = 2 s

B) the speed when reaching the ground

       v = v₀ - g t

       v = 0 - 9.8 2

       v = - 19.6 m / s

The negative sign indicates that the speed is down

In conclusion using the kinematic relations we can find the results for the different questions are:

   1.  The acceleraations is a = 375 m/s²

   2. The time is t = 62.5 s

   3. The velocity at the ground is  v = -14.7 m / s

   4.  The time to the bottoncliff is    1.7 s

   5.  The height isy₀ = 122.5 m

   6.

      A) The velocity is v = 25.2 m / s

      B)  The height  for t=1 s is  y = 18.1 m

  7)

      A)  Time in the air is t = 2 s

      B)  Te final veloicty is   v = - 19.6 m / s

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