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
