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tangare [24]
3 years ago
8

Kamal said the distance from the top of the balloon to the ground in the Example is Ï···· 353 ft. What mistake might Kamal have

made? 1 2
Physics
1 answer:
snow_lady [41]3 years ago
7 0

Complete question is;

Kamal said the distance from the top of the balloon to the ground in the Example image attached is √353 ft. What mistake might Kamal have made?

Answer:

the mistake Kamal made is that she probably used 17 ft as the perpendicular side of the triangle with b as the hypotenuse instead of using 17ft as the hypotenuse

Explanation:

From the image attached, we can see that the distance from the top of the balloon which is blue in color to the ground is denoted by "b".

Now the triangle is a right angle triangle with hypotenuse = 15ft + 2ft = 17 ft; the adjacent side = 8 ft, while the opposite side is "b".

Thus, we can use pythagoras theorem to solve this as;

b = √(17² - 8²)

b = √(289 - 64)

b = √225

b = 15ft

However,we are told Kamal got b as √353 ft.

From inspection of the calculations we just did, if we had used addition instead of subtraction, we would have gotten b = √353 ft.

Thus, we can under that the mistake Kamal made is that she probably used 17 ft as the perpendicular side of the triangle with b as the hypotenuse instead of using 17ft as the hypotenuse.

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An airplane is flying at a speed of 200 m/s in level flight at an altitude of 800 m. A package is to be dropped from the airplan
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2560m or 2.56km (rounded to 3 significant figures)

Explanation:

First, list all known and desired values/variables (initial vertical velocity is 0 as the plane is kept level and vertical acceleration is just gravity):

Vertical \ velocity \ (\frac{m}{s} ) =  u_{v} = 0 \\\\ Horizontal \ velocity \ (\frac{m}{s} ) =  u_{h} = 200\\\\ Vertical \ acceleration \ (\frac{m}{s^{2} } ) =  a_{v} =  9.8 \\\\ Horizontal \ acceleration \ (\frac{m}{s^{2} } ) =  a_{h} =  0 \\\\ Vertical \ displacement \ (m) = s_{v} = 800 \\\\ Horizontal \ displacement \ (m) = s_{h}

The horizontal displacement is going to be the distance travelled, horizontally of course, once the package is released;

First thing to understand is that the vertical and horizontal components are to be dealt with separately because they don't affect each other;

Since there is no horizontal acceleration (ignoring air resistance), we simply require a velocity and time to find the horizontal displacement, using the formula v = d/t (or speed = distance/time);

What we have is the horizontal velocity but we don't have the time taken;

One thing we know is that the time elapsed for the vertical fall of 800m and for the horizontal displacement must be the same;

What we do, therefore, is find the time taken for the vertical displacement using the formula, s = ut + ¹/₂·at², since we know the vertical velocity, height and acceleration:

800 = (0)t + ¹/₂·(9.8)t²

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t = 12.77...

We now have the time taken for the vertical fall and the horizontal displacement, we can use this with the horizontal velocity we know already and get the horizontal displacement:

u_{h} = \frac{s_{h} }{t} \\\\ 200 = \frac{s_{h} }{12.77...} \\\\ s_{h} = 200(12.77...) \\\\ s_{h} = 2555.5...

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