A helicopter which was ascending vertically at a steady velocity of 20 m/s, released a parcel that took 20 seconds to reach the ground. i) Calculate the time taken by the parcel to reach the maximum height.
The correct answer and explanation is:
To solve this problem, we need to break it into two parts: the upward motion of the parcel and its downward motion. Here’s how we can approach it:
Given Information:
- Initial velocity of the helicopter (and the parcel as it is released) = 20 m/s (upward).
- Time taken for the parcel to hit the ground = 20 seconds.
- Acceleration due to gravity, gg = 9.8 m/s² (downward).
Step 1: Time to Reach Maximum Height
When the parcel is released, it initially has an upward velocity of 20 m/s. As it ascends, the velocity gradually decreases due to gravity, until it reaches its maximum height where the velocity becomes zero. The time it takes for the parcel to reach maximum height can be found using the following kinematic equation: v=u+atv = u + at
where:
- vv is the final velocity (0 m/s at maximum height),
- uu is the initial velocity (20 m/s),
- aa is the acceleration due to gravity (-9.8 m/s²),
- tt is the time to reach maximum height (which we need to find).
Substituting the values: 0=20+(−9.8)t0 = 20 + (-9.8) t t=209.8≈2.04 secondst = \frac{20}{9.8} \approx 2.04 \, \text{seconds}
So, the time taken to reach the maximum height is approximately 2.04 seconds.
Step 2: Time to Fall Back to the Ground
Since the total time from release to hitting the ground is 20 seconds, and the time to reach the maximum height is 2.04 seconds, the time taken for the parcel to fall back down is: Time to fall=20−2.04=17.96 seconds\text{Time to fall} = 20 – 2.04 = 17.96 \, \text{seconds}
Conclusion:
The time taken by the parcel to reach the maximum height is approximately 2.04 seconds. The remaining time (17.96 seconds) is the time it takes for the parcel to fall back to the ground, where the upward motion is countered by gravity before it accelerates downward.