Question

I NEED HELP I AM SO CONFUSED, WILL GIVE BRAIN.

Answer 1

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 = $\frac{v-v_o}{t}$

             a = $\frac{17500-10000}{20}$

             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 = $\sqrt{\frac{2x}{a} }$

             t = $\sqrt{\frac{2 \ 125}{4} }$

             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 = $\sqrt{\frac{2y_o}{g} }$

       t = $\sqrt{\frac{2 \ 20}{9.8} }$

       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

Learn more here: brainly.com/question/15068914