Wes and Lindsay stand on the roof of a building. Wes leans over the edge and drops an apple. Lindsay waits, 1.25s afrer Wes releases his fruit and throws an orange straight down at 28 m/s. Both pieces of fruit hit the ground simultaneously. Calculate the common height from which the fruit were released. Ignore the effects of air resistence.
Bob, Im not getting the same answer when i do plug in the H into finding the height of the apples and oranges
To solve this problem, we can use the kinematic equation for displacement:
\[y = v_0t + \frac{1}{2}at^2\]
where
- \(y\) is the height or displacement
- \(v_0\) is the initial velocity
- \(t\) is the time
- \(a\) is the acceleration
Since both fruit hit the ground simultaneously, their time of flight must be the same.
First, let's calculate the time it takes for Wes' apple to fall to the ground. We know that the initial velocity (\(v_0\)) is 0 m/s, as Wes dropped the apple from rest. The acceleration (\(a\)) due to gravity is approximately -9.8 m/s² (taking downward as negative). Let's assume that the height from which the fruit were released is \(h\).
Using the equation for displacement, we can set \(y\) as -h (taking downward as negative):
\[-h = 0 \times t + \frac{1}{2}(-9.8)t^2\]
Simplifying the equation:
\[-h = -4.9t^2\] (Equation 1)
Next, let's calculate the time it takes for Lindsay's orange to fall to the ground. We know that the initial velocity (\(v_0\)) is 28 m/s (thrown straight down). The acceleration (\(a\)) due to gravity is still -9.8 m/s². Again, let's assume the height from which the fruit were released is \(h\).
Using the equation for displacement, we can set \(y\) as -h:
\[-h = 28 \times t + \frac{1}{2}(-9.8)t^2\]
Simplifying the equation:
\[-h = 28t - 4.9t^2\] (Equation 2)
Now, we can equate Equation 1 and Equation 2, as the two times of flight are equal:
\[-4.9t^2 = 28t - 4.9t^2\]
Simplifying the equation:
\[28t - 4.9t^2 = 0\]
Dividing both sides by \(t\) (because \(t\) cannot be 0):
\[28 - 4.9t = 0\]
Solving for \(t\):
\[4.9t = 28\]
\[t = \frac{28}{4.9}\]
\[t \approx 5.71\]
Now that we have the time it takes for the fruit to fall, we can substitute this value into Equation 1 to solve for \(h\):
\[-h = -4.9 \times (5.71)^2\]
Simplifying the equation:
\[h = 4.9 \times (5.71)^2\]
Calculating the value:
\[h \approx 162.25\]
Therefore, the common height from which the fruit were released is approximately 162.25 meters.
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.
the height is the same, as is the time (one is 1.25 seconds longer)
happle=1/2 g t^2
horante=-28*(t-1.25)+1/2 g (t-1.25)^2
solve for t in the first equation
t=sqrt(2h/g)
put that t into the second equation
h=-28(sqrt(2h/g) - 1.25)-4.9(sqrt (2h/g)-1.25)^2
expand this out, the height h divides out, solve for t. Then, calculate h.