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BartSMP [9]
3 years ago
11

A weather balloon, rising vertically at 25.0 m/s, releases an instrument package when it is 100. meters above the ground. How ma

ny seconds does it take from when the package was released for the package to hit the ground?
Physics
1 answer:
Illusion [34]3 years ago
4 0

Answer:

7.74 seconds

Explanation:

u = 25 m/s

h = - 100 m

Let it takes time t to hit the ground.

Use second equation of motion

h = ut-\frac{1}{2}gt^{2}

-100 = 25t-4.9t^{2}

4.9t^{2}-25t-100=0

t=\frac{25\pm \sqrt{625+4\times 4.9\times 100}}{2\times4.9}

t=\frac{25\pm 50.84}{9.8}

Take positive sign as time can never be negative.

t = 7.74 s

Thus, the package takes 7.74 seconds to reach the ground.

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A shot-putter projects the shot at 42.00˚ to the horizontal from a height of 2.100 m. It lands 17.00 m away horizontally. Next,
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Answer:

Explanation:

Let 100 m/s  be the velocity of projection.

So horizontal component

= 100 cos42

= 74.31 m /s

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Net displacement = 2.1 downwards ( + ve )

Using s = ut + 1/2 gt²

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If angle of projction is 40°

So horizontal component

= 100 cos40

= 76.60 m /s

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Net displacement = 2.1 downwards ( + ve )

Using s = ut + 1/2 gt²

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So horizontal range is increased , if angle of projection is increased .

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