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3 years ago

14

If a box of supplies is dropped from the cargo hold of an airplane travelling at an altitude of 4,410 meters how long will it ta

ke to reach the ground!

1 answer:

Paraphin [41]3 years ago

5 0

Answer:

It will take 30 seconds to reach the ground, and it will be travelling at 294 m/s when it does so. This means that its average velocity was 147 m/s.

Explanation:

$d=v_ot+\dfrac{1}{2}at^2$

Since the initial velocity of a dropped object is 0, we can make this the equation:

$d=\dfrac{1}{2}at^2 \\\\4410=\dfac{1}{2}(9.8)t^2 \\\\t^2=900$

$t=30\text{ seconds}$

The final velocity can be calculated with the formula:

$v_f=v_o+at$

Once again, since there is no initial velocity:

$v_f=at \\\\v_f=(9.8)(30)=294m/s$

Since the initial velocity is 0, the average vertical velocity is 294/2=147 m/s.

Hope this helps!

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A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10 m/s. The instan

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Answer:

The velocity just before hitting the ground is $v_f = 30 m/s$

Explanation:

From the question we are told that

The initial speed is $u = 10 m/s$

The final speed is $v = 30 \ m/s$

From the equations of motion we have that

$v^2 =u^2 + 2as$

Where s is the distance travelled which is the height of the cliff

So making it the subject of the the formula we have that

$s = \frac{v^2 - u^2 }{2a}$

Where a is the acceleration due to gravity with a value $a = 9.8m/s^2$

So

$s = \frac{30^2 - 10^2 }{2 * 9.8 }$

$s = 40.8 \ m$

Now we are told that was through horizontally with a speed of

$v_x =10 m/s$

Which implies that this would be its velocity horizontally through out the motion

Now it final velocity vertically can be mathematically evaluated as

$v_y = \sqrt{2as}$

Substituting values

$v_y = \sqrt{(2 * 9.8 * 40.8)}$

$v_y = 28.3 \ m/s$

The resultant final velocity is mathematically evaluated as

$v_f = \sqrt{v_x^2 + v_y^2}$

Substituting values

$v_f = \sqrt{10^2 + 28.3^2}$

$v_f = 30 m/s$

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A high fountain of water is located at the center of a circular pool. Not wishing to get his feet wet, a student walks around th

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Answer:

The fountain is 3.43 m high.

Explanation:

Circumference of the pool = 15 m.

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r = 2.3864

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Tan θ = $\frac{opposite}{adjacent}$

Tan 55 = $\frac{h}{2.4}$

h = Tan 55 x 2.4

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