A helicopter flying horizontally at a velocity of 25m/s drops a mailbag from a height of 15m to a letter carrier waiting on the ground below.a) how long will it take the bag to fall the ground?b) how far in advance of the letter carrier must the bag be released so that it lands at her feet?
Answer (1)
Answer:To solve this problem, we'll break it down into two parts: the time it takes for the mailbag to fall and the horizontal distance it travels during that time.### a) Time to fall to the groundWe can use the following equation of motion for free fall:\[h = \frac{1}{2}gt^2\]where:- \( h \) is the height (15 m),- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),- \( t \) is the time in seconds.Rearranging the formula to solve for \( t \):\[t^2 = \frac{2h}{g}\]Substituting the values:\[t^2 = \frac{2 \times 15}{9.81}\]\[t^2 \approx \frac{30}{9.81} \approx 3.058\]Taking the square root to find \( t \):\[t \approx \sqrt{3.058} \approx 1.74 \, \text{s}\]### b) Horizontal distance traveledNow that we have the time, we can calculate how far the bag will travel horizontally during that time. The horizontal distance \( d \) can be found using:\[d = v \cdot t\]where \( v \) is the horizontal velocity (25 m/s).Substituting the values:\[d = 25 \, \text{m/s} \times 1.74 \, \text{s} \approx 43.5 \, \text{m}\]### Summary of Results- **a)** The time it takes for the bag to fall to the ground is approximately **1.74 seconds**.- **b)** The bag must be released approximately **43.5 meters** in advance of the letter carrier to land at her feet.
