To solve this problem, we can use the equations of motion for each piece of fruit and equate their times of flight. The equations of motion for vertical motion under gravity are:
For the apple:
h_a = h_0 + v_0t - (1/2)gt^2
For the orange:
h_o = h_0 + v_0t - (1/2)gt^2
Where:
h_a and h_o are the heights of the apple and orange from the ground,
h_0 is the initial height from which the fruits were released (common height we want to find),
v_0 is the initial velocity of the orange,
t is the time of flight, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the apple was dropped, its initial velocity v_0 is zero, whereas the orange has an initial velocity of 28 m/s.
Now, we can solve for h_0 by setting up and equating the equations of motion for the apple and the orange. We know that the duration for the orange is 1.25 seconds longer than that for the apple (t_a + 1.25s = t_o).
For the apple:
h_a = h_0 + 0 - (1/2)gt_a^2
For the orange:
h_o = h_0 + 28t_o - (1/2)gt_o^2
Using the relation t_a + 1.25s = t_o, we can substitute t_o - 1.25s in place of t_a in the equation for the apple:
h_a = h_0 + 0 - (1/2)g(t_o - 1.25s)^2
Now we can equate h_a and h_o:
h_0 + 0 - (1/2)g(t_o - 1.25s)^2 = h_0 + 28t_o - (1/2)gt_o^2
Simplifying the equation, we get:
0 - (1/2)g(t_o - 1.25s)^2 = 28t_o - (1/2)gt_o^2
Expanding the squared term and rearranging the equation, we get:
(1/2)gt_o^2 - 28t_o - (1/2)g(t_o - 1.25s)^2 = 0
This quadratic equation in terms of t_o can be solved using the quadratic formula:
t_o = [-(-28) ± √((-28)^2 - 4(1/2)g(-(1/2)g(t_o - 1.25s)^2)) / (2(1/2)g)]
Simplifying further, we have:
t_o = [28 ± √(784 - g(t_o - 1.25s)^2) / g]
To solve, we can make an initial guess for t_o and iteratively refine it using this equation until we converge on a value for t_o.
Once we have t_o, we can substitute it back into either the equation for the apple or the orange to find the common height h_0 from which the fruit were released.