Respuesta :
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
The student determine the actual speed of the object when it reaches the ground as 12.52 m/s.
Given data:
The distance from the Earth's surface is, = 10 m.
Time taken to reach the ground is, t = 1.43 s.
The speed of object is, v = 14.3 m/s.
Experimental value of time interval is, t' = 3.2 s.
Use kinematic equation of motion to compute true value for acceleration of the ball as it reaches the ground:
[tex]h=ut+\dfrac{1}{2}a't'^{2} \\\\10=0 \times t+\dfrac{1}{2} \times a' \times 3.2^{2} \\\\a'=\dfrac{20}{3.2^{2}}\\\\a'= 1.95 \;\rm m/s^{2}[/tex]
Now, use the principle of conservation of total energy of system:
Potential energy - work done by air resistance = Kinetic energy
[tex]mgh-(ma) \times h=\dfrac{1}{2}mv^{2} \\\\gh-(a) \times h=\dfrac{1}{2}v^{2} \\\\v=\sqrt{2h(g-a)}[/tex]
Here, v is the actual speed of object while reaching the ground.
Solving as,
[tex]v=\sqrt{2 \times 10(9.8-1.95)}\\\\v=12.52 \;\rm m/s[/tex]
Thus, we can conclude that the student determine the actual speed of the object when it reaches the ground as 12.52 m/s.
Learn more about the conservation of energy here:
https://brainly.com/question/2137260
