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GalinKa [24]
3 years ago
11

A life preserver is thrown from an helicopter straight down to a person in distress. The initial velocity of the life preserver

is 1.60 m/s and it takes 2.3 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note Ignore air resistance.
Physics
1 answer:
Leno4ka [110]3 years ago
6 0

Answer:29.627 m

Explanation:

Given

Initial velocity of life preserver(u) is 1.6 m/s

it takes 2.3 s to reach the water

using equation of motion

v=u+at

v=1.6+9.81\times 2.3

v=24.163 m/s

Let s be the height of life preserver

v^2-u^2=2gs

24.163^2-1.6^2=2\times 9.81\times s

s=\frac{581.29}{2\times 9.81}

s=29.627 m

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Answer:

FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

Explanation:

First you have to consider that the Ford Thunderbird (FT) follows a rectilinear motion with varying acceleration, while Mercedez Benz (MB) has a constant velocity (no acceleration). So if you finde the time spent by FT in each section, and the distance, then you will find the distance for MB.

1) Vf² = Vi² + 2ad, where Vf: final velocity, Vi: ionitial velocity, a: acceleration and d: distance.

For the first portion  (0 m/s)² = (78.5 m/s)² + 2a(250 m) ⇒

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Now, you can find the corresponding time for this section with the following formule: Vf = Vi + at ⇒ 0 m/s = 78.5 m/s + (-12.3 m/s²) t

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2) Then FT spent 5 seconds in the pit.

3) The the FT accelerates until reach 78.5 m/s again in a distance of 370 m.

Vf² = Vi² + 2ad ⇒ (78.5 m/s)² = (0 m/s)² + 2a(370 m)

⇒ (78.5 m/s)²/ 2(370 m) = a ⇒ a = 8.3 m/s²

Then, Vf = Vi + at ⇒ 78.5 m/s = 0 m/2 + (8.3 m/s²) t

⇒ (78.5 m/s)/(8.3 m/s²) = t ⇒ t = 9.5 seconds.

4) Summarizing, the FT moves 620 meters (250 + 370 mts) in 20.9 seconds ( 6.4 s + 5 s + 9.5 s).

5) During this time, MB moves

Velocity = distance/ time ⇒ Velocity x time = Distance

⇒ Distance = (78.5 m/s) x  (20.9 seconds) ⇒ Distance = 1640.6 meters

6) Finally, the FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

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