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zmey [24]
3 years ago
10

A package is dropped from a helicopter moving upward at 15m/s. If it takes 10 s before the package strikes the ground, how high

above the ground was the ground was the package when it was released if air resistance is negligible?
(A) 408 m
(B) 272 m
(C) 204 m
(D) 340 m
Physics
2 answers:
Vlad1618 [11]3 years ago
6 0

Answer:

A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released? (Disregard air resistance.)

Show step by step please.

Note: The answer is given it's should be 1000 m ??

This what i can up with so see what it is kid

Explanation:

Anni [7]3 years ago
4 0

Answer:

The right answer is (D) 340m

Explanation:

If the package is dropped from a helicopter moving upward and the air resistance is negligible, the only acting force in the package is the gravity force. Therefore you can consider this as a Vertical Projectile Motion problem.

The initial speed of the package V₀ is 15m/s.

The acceleration of the package G is 9.8m/s².

The initial hight is Y₀. Therefore:

Y(t)=Y₀+V₀·t-0.5·G·t²

But we know than T(10s)=0m

0m=Y₀+150m-490m

Y₀=340m

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g A ball thrown straight up into the air is found to be moving at 7.94 m/s after falling 2.72 m below its release point. Find th
kati45 [8]

The ball has height <em>y</em> and velocity <em>v</em> at time <em>t</em> according to

<em>y</em> = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

and

<em>v</em> = <em>v</em>₀ - <em>g t</em>

where <em>v</em>₀ is its initial speed and <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time <em>t </em>such that

-2.72 m = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

-7.94 m/s = <em>v</em>₀ - <em>g t</em>

Solve for <em>t</em> in the second equation:

<em>t </em>= (<em>v</em>₀ + 7.94 m/s)/<em>g</em>

Substitute this into the first equation and solve for <em>v</em>₀ :

-2.72 m = <em>v</em>₀ (<em>v</em>₀ + 7.94 m/s) /<em>g</em> - 1/2 <em>g</em> ((<em>v</em>₀ + 7.94 m/s)/<em>g</em>)²

-2.72 m = <em>v</em>₀²/<em>g</em> + (7.94 m/s) <em>v</em>₀/<em>g</em> - 1/2 (<em>v</em>₀ + 7.94 m/s)²/<em>g</em>

2 (-2.72 m) <em>g</em> = 2<em>v</em>₀² + 2 (7.94 m/s) <em>v</em>₀ - (<em>v</em>₀ + 7.94 m/s)²

2 (-2.72 m) (9.80 m/s²) = 2<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ - (<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ + 63.0 m²/s²)

-53.3 m²/s² = <em>v</em>₀² - 63.0 m²/s²

<em>v</em>₀² = 9.73 m²/s²

<em>v</em>₀ = 3.12 m/s

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