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Sagot :
To find the maximum height that a stone will reach when it is thrown upwards with an initial velocity [tex]\(v = 20 \, \text{m/s}\)[/tex] and [tex]\(g = 10 \, \text{m/s}^2\)[/tex], we can use the following formula from kinematics:
[tex]\[ h = \frac{v^2}{2g} \][/tex]
Here,
- [tex]\(v\)[/tex] is the initial velocity,
- [tex]\(g\)[/tex] is the acceleration due to gravity, and it is given that [tex]\(g = 10 \, \text{m/s}^2\)[/tex].
Let's plug in the values into the formula:
[tex]\[ h = \frac{(20 \, \text{m/s})^2}{2 \times 10 \, \text{m/s}^2} \][/tex]
First, calculate the square of the initial velocity:
[tex]\[ (20 \, \text{m/s})^2 = 400 \, \text{m}^2/\text{s}^2 \][/tex]
Next, multiply the gravity by 2:
[tex]\[ 2 \times 10 \, \text{m/s}^2 = 20 \, \text{m/s}^2 \][/tex]
Now, divide the square of the initial velocity by this product:
[tex]\[ h = \frac{400 \, \text{m}^2/\text{s}^2}{20 \, \text{m/s}^2} = 20 \, \text{m} \][/tex]
Thus, the maximum height the stone will reach is [tex]\(20 \, \text{meters}\)[/tex].
Therefore, the correct answer is [tex]\( B: 20 \, \text{m} (\text{20 मी}) \)[/tex].
[tex]\[ h = \frac{v^2}{2g} \][/tex]
Here,
- [tex]\(v\)[/tex] is the initial velocity,
- [tex]\(g\)[/tex] is the acceleration due to gravity, and it is given that [tex]\(g = 10 \, \text{m/s}^2\)[/tex].
Let's plug in the values into the formula:
[tex]\[ h = \frac{(20 \, \text{m/s})^2}{2 \times 10 \, \text{m/s}^2} \][/tex]
First, calculate the square of the initial velocity:
[tex]\[ (20 \, \text{m/s})^2 = 400 \, \text{m}^2/\text{s}^2 \][/tex]
Next, multiply the gravity by 2:
[tex]\[ 2 \times 10 \, \text{m/s}^2 = 20 \, \text{m/s}^2 \][/tex]
Now, divide the square of the initial velocity by this product:
[tex]\[ h = \frac{400 \, \text{m}^2/\text{s}^2}{20 \, \text{m/s}^2} = 20 \, \text{m} \][/tex]
Thus, the maximum height the stone will reach is [tex]\(20 \, \text{meters}\)[/tex].
Therefore, the correct answer is [tex]\( B: 20 \, \text{m} (\text{20 मी}) \)[/tex].
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