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.
This is a duplicate post. Please don't clutter the website with identical posts
That method will give the wrong answer unless you use the kinetic energy change when it hits the ground.
I saw that after I did, I should have had him tackle it with the kinematic equations, but I believe you corrected this in a later post. Consequently, I didn't bother to post again saying ignore the setup.
To solve this problem, we need to first determine the time it takes for the apple to fall from the height of the building and the time it takes for the orange to fall from the same height. Since both pieces of fruit hit the ground simultaneously, their fall times will be equal.
We can use the equation for the distance traveled by a falling object without air resistance:
d = (1/2) * g * t^2
Where:
d is the distance traveled
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time taken
For the apple:
d_apple = (1/2) * g * t_apple^2
For the orange:
d_orange = (1/2) * g * t_orange^2
Since the orange is thrown 1.25 seconds after the apple is dropped, the time for the apple is t_apple and the time for the orange is (t_apple + 1.25).
Therefore:
d_apple = (1/2) * g * t_apple^2
d_orange = (1/2) * g * (t_apple + 1.25)^2
Since both pieces of fruit are dropped from the same height, their distances traveled will be the same:
d_apple = d_orange
(1/2) * g * t_apple^2 = (1/2) * g * (t_apple + 1.25)^2
Now, we can solve this equation to find t_apple and then calculate the height of the building.
Let's rearrange the equation to solve for t_apple:
t_apple^2 = (t_apple + 1.25)^2
Expand the equation:
t_apple^2 = t_apple^2 + 2 * t_apple * 1.25 + 1.25^2
Simplify the equation:
0 = 2 * t_apple * 1.25 + 1.25^2
Now, solve for t_apple:
2 * t_apple * 1.25 = -1.25^2
Divide both sides by 2 * 1.25:
t_apple = (-1.25^2) / (2 * 1.25)
Calculate t_apple:
t_apple = (-1.5625) / 2
t_apple = -0.78125 s
Since time cannot be negative, this result is not valid. This means that the equation (1/2) * g * t_apple^2 = (1/2) * g * (t_apple + 1.25)^2 cannot be satisfied, indicating that there might be an error or inconsistency in the given information or calculations.
Please double-check the given problem or provide additional information if necessary to further resolve the issue and find the common height from which the fruits were released.
I would just do this by looking at Lindsay's orange, and use conservation of kinetic energy to solve this problem. KE=PE=1/2(mv^2)=mgh and solve for h.
g=9.8m/s^2
1/2(mv^2)=mgh
1/2(v^2)=gh
The masses are the same so they cancel out.
(1/2(28m/s)^2/9.8m/s^2)=h