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.
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To solve this problem, we can use the equations of motion to determine the height from which the fruit were released.
We'll start by considering the motion of Wes's apple. We know that the only force acting on it is gravity, so it will undergo free fall motion. The equation for the distance traveled by an object in free fall is given by:
d = (1/2) * g * t^2
where d is the distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken.
Since Wes drops the apple, the time taken for it to hit the ground is 1.25 seconds. Substituting the known values into the equation, we get:
d = (1/2) * 9.8 * (1.25)^2
d = 7.34375 m
Now let's consider the motion of Lindsay's orange. It is thrown downwards with an initial velocity of 28 m/s. The equation for the distance covered by an object under constant acceleration is given by:
d = v * t + (1/2) * a * t^2
where v is the initial velocity, t is the time taken, and a is the acceleration.
In this case, the acceleration is also due to gravity, but since the orange is thrown downwards, the acceleration will be negative (-9.8 m/s^2). Again, substituting the known values into the equation, we get:
d = 28 * 1.25 + (1/2) * (-9.8) * (1.25)^2
d = 17.5 - 7.34375
d = 10.15625 m
Therefore, the common height from which the fruit were released is 10.15625 meters.