A small parachute is dropped from a cliff. it falls at a constant speed rate 2 m/s. A small rock is dropped 20.0 seconds later. how far will the rock drop before it passes the parachute?
We want the distances fallen to be equal. The parachute falls for 20 seconds more than the rock:
2(t+20) = 4.9t^2
4.9t^2 - 2t - 40 = 0
t = 3.07
so, after about 3 seconds, the rock either passes the chute, or it hits it.
To find out how far the rock will drop before it passes the parachute, we need to determine the distance the parachute has dropped during the 20.0 seconds.
Since the parachute falls at a constant speed of 2 m/s, we can calculate the distance it has dropped using the formula:
Distance = Speed × Time
Substituting the values into the formula:
Distance = 2 m/s × 20.0 s
Distance = 40.0 meters
Therefore, the parachute will have dropped 40.0 meters during the 20.0 seconds.
Now, let's calculate the distance the rock will drop before it passes the parachute. Since the rock is dropped 20.0 seconds after the parachute, we only need to consider the time it takes for the rock to catch up with the parachute.
Since both the rock and the parachute fall at the same rate (gravity), the rock will need to close the initial 40.0 meters gap between them.
We can use the formula for distance dropped under constant acceleration:
Distance = 0.5 × Acceleration × Time²
Assuming that acceleration due to gravity is 9.8 m/s², we can substitute the values into the formula:
Distance = 0.5 × 9.8 m/s² × (Time + 20.0 s)²
Calculating this, we have:
Distance = 0.5 × 9.8 m/s² × (20.0 s)²
Distance = 0.5 × 9.8 m/s² × 400.0 s²
Distance = 0.5 × 9.8 m/s² × 1600.0 m²
Distance = 0.5 × 980.0 m²
Distance = 490.0 meters
Therefore, the rock will drop 490.0 meters before it passes the parachute.