posted by Wes
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.