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A stone is thrown vertically upwards with a speed of 20 m/s. How high will it go before it begins to fall? (g = 9.8 m/s²)​

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User Kalamar Obliwy
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Answer:

20.41 meters

Step-by-step explanation:

We can solve this problem using the equations of motion.

When the stone is thrown upwards, its initial velocity (u) is +20 m/s (positive because it's in the upward direction) and its acceleration (a) is -9.8 m/s² (negative because the acceleration due to gravity is in the downward direction). We want to find the maximum height (h) that the stone reaches before it begins to fall back down.

The equation that relates these quantities is:

v² = u² + 2as

where v is the final velocity (which is zero when the stone reaches its maximum height), s is the displacement (which is equal to the maximum height we want to find), and the other variables are as described above.

Solving for s, we get:

s = (v² - u²) / 2a

Since v is zero, we can simplify this to:

s = u² / 2a

Substituting the given values, we get:

s = (20 m/s)² / (2 x -9.8 m/s²) = 20.41 m

Therefore, the stone will reach a maximum height of 20.41 meters before it begins to fall back down.

User Pugsley
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