Probe speed A data collection probe is dropped from a stationa...

Question

Probe speed A data collection probe is dropped from a stationary balloon, and it falls with a velocity (in m/s) given by v(t) = 9.8t, neglecting air resistance. After 10 s, a chute deploys and the probe immediately slows to a constant speed of 10 m/s, which it maintains until it enters the ocean.

c. If the probe was released from an altitude of 3 km, when does it enter the ocean?

c. If the probe was released from an altitude of 3 km, when does it enter the ocean?

Accepted Answer

Welcome back, everyone. A package is dropped from a stationary drone in full sweat velocity in meters per second given by V of T equals 7.2 T, neglecting air resistance. After 8.0 seconds, a safety device activates and the package slows instantly to a constant speed of 9.0 m per second, which it maintains until it lands. If the package is dropped from 1800 m, in how many seconds after the release does it land round to the nearest integer? A 134 seconds B 142 seconds. C 182 seconds, and D 206 seconds. So first of all, we have to understand that our package. is thrown from a stationary thrown with a velocity of 7.2 T. It is a velocity function, right? So, what we can do is simply identify the displacement for the first segment, let's call it D1. And let's recall that displacement is equal to a definite integral of the velocity function, right V of TDT between two time values. Our initial time is 0, and our velocity function holds up to 8 seconds. So we're going to integrate from 0 to 8, and we can replace V of T with the actual expression of the velocity function. So that's 7.2 C, right? We're going to integrate with respect to time. We can factor out the constant 7.2. The integral of T is T squared divided by 2, and we want to evaluate the result from 0 to 8. Simplifying, we get 3.6 T squared evaluated from 0 to 8. And this is equal to 3.682 minus 0 squad. Let's calculate the result. We got 230.4 m. And now what we can do is simply find the remaining distance to fall, right? So, our next segment D2 is going to be our total height D minus D1. Which would be 1800 m minus 230.4 m. The remaining distance is equal to 1,569.6 m. And we know that our package falls at a constant velocity, right? So what we can do is simply find the required time value, let's call it T2, to fall that distance. Let's recall the distance formula. Distance equals velocity multiplied by time if our velocity is constant. So time is the ratio between distance and velocity. So we're going to take D2 and divide by V2. Our D2 is 1,569.6 m. And we know that our constant velocity is 9.0 m per second. Dividing, we get 174.4 seconds. And now we can identify the total time of flight. The total time T is Our first time segment plus our second time segment. Our first segment consists of 8.0 seconds, and we are going to add 174.4 seconds identified previously. We get 182.4 seconds, which is approximately 182 seconds. This corresponds to the answer choice C. Thank you for watching.

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