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.
Start counting time when the apple is thrown. The distance it falls is
D1 = (g/2)t^2 = 4.9 t^2
For the orange, for t>1.25 s,
D2 = 28(t-1.25) + 4.9(t-1.25)^2
Set D1 = D2 and solve for t when they are both at the same distance (ground level).
4.9t^2 = 28(t-1.25) + 4.9(t-1.25)^2
= 28(t-1.25) + 4.9t^2 -12.25 t + 7.656
0 = 15.75 t - 27.34
t = 1.736 s
Use that t in the D1 formula for the building height.
I get 14.8 meters. Check my math.
To solve this problem, we can use the equations of motion under constant acceleration. Let's define a few variables:
- Let h be the common height from which the fruit were released.
- Let t1 be the time it takes for Wes' apple to reach the ground.
- Let t2 be the time it takes for Lindsay's orange to reach the ground.
We can start by finding the equations of motion for each fruit using the formula h = vit + 0.5at^2, where:
- h is the vertical displacement (the height)
- vi is the initial velocity
- a is the acceleration (due to gravity, which is approximately 9.8 m/s^2 near the Earth's surface)
- t is the time
For Wes' apple:
The initial velocity is vi = 0 (since it's simply dropped)
The equation becomes h1 = 0 + 0.5 * 9.8 * t1^2, which simplifies to h1 = 4.9 * t1^2.
For Lindsay's orange:
The initial velocity is vi = 28 m/s (she throws it straight down)
The equation becomes h2 = 28 * t2 + 0.5 * 9.8 * t2^2, which simplifies to h2 = 4.9 * t2^2 + 28 * t2.
Since both fruits hit the ground simultaneously, their times of flight are equal, so we have t1 = t2 + 1.25s (since Lindsay threw her orange 1.25s after Wes dropped his apple).
Now we can solve the equations by setting them equal to each other:
4.9 * t1^2 = 4.9 * t2^2 + 28 * t2
Substituting t2 + 1.25 for t1:
4.9 * (t2 + 1.25)^2 = 4.9 * t2^2 + 28 * t2
Expanding and simplifying the equation:
4.9 * (t2^2 + 2.5 * t2 + 1.5625) = 4.9 * t2^2 + 28 * t2
Now we can solve for t2:
12.25t2 + 6.0625 = 28t2
15.75t2 = 6.0625
t2 = 0.3854s (approximately)
Now we can plug t2 back into one of the original equations to find h:
h2 = 4.9 * t2^2 + 28 * t2
h2 = 4.9 * (0.3854)^2 + 28 * 0.3854
h2 ≈ 0.7461 + 10.7972
h2 ≈ 11.5433m
Therefore, the common height from which the fruit were released is approximately 11.5433 meters.