Daniyal Ahmad
Jul 17
A body falls from a height h=200m . The ratio of distance travell in each 2 sec t=0 to t=6 is?
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ChatGPT
Jul 17
To find the ratio of distance traveled in each 2-second interval, we first need to calculate the total time it takes for the body to fall from a height of 200m.
Using the formula for free fall distance:
h = (1/2)gt^2
200 = (1/2)(9.8)t^2
t^2 = 40.82
t ≈ 6.39 seconds
Since the total time taken is approximately 6.39 seconds, we can divide this time into 3 intervals of 2 seconds each.
Distance traveled in the first 2 seconds:
h1 = (1/2)gt^2
h1 = (1/2)(9.8)(2)^2
h1 = 19.6m
Distance traveled in the next 2 seconds:
h2 = (1/2)g(t+2)^2 - (1/2)gt^2
h2 = (1/2)(9.8)(4)^2 - (1/2)(9.8)(2)^2
h2 = 78.4m - 19.6m
h2 = 58.8m
Distance traveled in the last 2 seconds:
h3 = h - h1 - h2
h3 = 200m - 19.6m - 58.8m
h3 = 121.6m
Therefore, the ratio of distance traveled in each 2-second interval is:
h1 : h2 : h3 = 19.6m : 58.8m : 121.6m
h1 : h2 : h3 = 1 : 3 : 6.2
Using the formula for free fall distance:
h = (1/2)gt^2
200 = (1/2)(9.8)t^2
t^2 = 40.82
t ≈ 6.39 seconds
Since the total time taken is approximately 6.39 seconds, we can divide this time into 3 intervals of 2 seconds each.
Distance traveled in the first 2 seconds:
h1 = (1/2)gt^2
h1 = (1/2)(9.8)(2)^2
h1 = 19.6m
Distance traveled in the next 2 seconds:
h2 = (1/2)g(t+2)^2 - (1/2)gt^2
h2 = (1/2)(9.8)(4)^2 - (1/2)(9.8)(2)^2
h2 = 78.4m - 19.6m
h2 = 58.8m
Distance traveled in the last 2 seconds:
h3 = h - h1 - h2
h3 = 200m - 19.6m - 58.8m
h3 = 121.6m
Therefore, the ratio of distance traveled in each 2-second interval is:
h1 : h2 : h3 = 19.6m : 58.8m : 121.6m
h1 : h2 : h3 = 1 : 3 : 6.2
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Google Gemini AI
Jul 17
The distance travelled by the body in the first 2 seconds is given by:
$$s_1 = h - \frac{1}{2}gt^2$$
$$s_1 = 200 - \frac{1}{2}\times 9.81 \times 2^2$$
$$s_1 = 200 - 19.62$$
$$s_1 = 180.38\ m$$
The distance travelled by the body in the next 2 seconds is given by:
$$s_2 = s_1 + h - \frac{1}{2}gt^2$$
$$s_2 = 180.38 + 200 - \frac{1}{2}\times 9.81 \times 4^2$$
$$s_2 = 180.38 + 200 - 78.48$$
$$s_2 = 301.9\ m$$
The distance travelled by the body in the third 2 seconds is given by:
$$s_3 = s_2 + h - \frac{1}{2}gt^2$$
$$s_3 = 301.9 + 200 - \frac{1}{2}\times 9.81 \times 6^2$$
$$s_3 = 301.9 + 200 - 176.58$$
$$s_3 = 425.32\ m$$
Therefore, the ratio of distance travelled in each 2 seconds is:
$$\frac{s_1}{s_2} = \frac{180.38}{301.9} = 0.597$$
$$\frac{s_2}{s_3} = \frac{301.9}{425.32} = 0.71$$.
$$s_1 = h - \frac{1}{2}gt^2$$
$$s_1 = 200 - \frac{1}{2}\times 9.81 \times 2^2$$
$$s_1 = 200 - 19.62$$
$$s_1 = 180.38\ m$$
The distance travelled by the body in the next 2 seconds is given by:
$$s_2 = s_1 + h - \frac{1}{2}gt^2$$
$$s_2 = 180.38 + 200 - \frac{1}{2}\times 9.81 \times 4^2$$
$$s_2 = 180.38 + 200 - 78.48$$
$$s_2 = 301.9\ m$$
The distance travelled by the body in the third 2 seconds is given by:
$$s_3 = s_2 + h - \frac{1}{2}gt^2$$
$$s_3 = 301.9 + 200 - \frac{1}{2}\times 9.81 \times 6^2$$
$$s_3 = 301.9 + 200 - 176.58$$
$$s_3 = 425.32\ m$$
Therefore, the ratio of distance travelled in each 2 seconds is:
$$\frac{s_1}{s_2} = \frac{180.38}{301.9} = 0.597$$
$$\frac{s_2}{s_3} = \frac{301.9}{425.32} = 0.71$$.
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