Question

An object is released from rest near and above Earth’s surface from a distance of 10m. After applying the appropriate kinematic equation, a student predicts that it will take 1.43s for the object to reach the ground with a speed of 14.3m/s . After performing the experiment, it is found that the object reaches the ground after a time of 3.2s. How should the student determine the actual speed of the object when it reaches the ground? Assume that the acceleration of the object is constant as it falls.

Answer 1

Answer:

v_y = 12.54 m/s

Explanation:

Given:

- Initial vertical distance y_o = 10 m

- Initial velocity v_y,o  = 0 m/s

- The acceleration of object in air = a_y

- The actual time taken to reach ground t = 3.2 s

Find:

- Determine the actual speed of the object when it reaches the ground?

Solution:

- Use kinematic equation of motion to compute true value for acceleration of the ball as it reaches the ground:

                             y = y_o + v_y,o*t + 0.5*a_y*t^2

                             0 = 10 + 0 + 0.5*a_y*(3.2)^2

                             a_y = - 20 / (3.2)^2 = 1.953125 m/s^2

- Use the principle of conservation of total energy of system:

                             E_p - W_f = E_k

Where,                  E_p = m*g*y_o

                             W_f = m*a_y*(y_i - y_f)      ..... Effects of air resistance

                             E_k = 0.5*m*v_y^2

Hence,                  m*g*y_o - m*a_y*(y_i - y_f) = 0.5*m*v_y^2

                             g*(10) - (1.953125)*(10) = 0.5*v_y^2

                             v_y = sqrt (157.1375)

                            v_y = 12.54 m/s