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//s^2)`

Answer

To solve the problem of how high a stone will go when thrown vertically upwards with an initial speed of 20 m/s, we can use the equations of motion. Here's the step-by-step solution:

### Step 1: Identify the known values
- Initial velocity (u) = 20 m/s (the speed with which the stone is thrown upwards)
- Final velocity (v) = 0 m/s (the speed at the highest point of its motion)
- Acceleration due to gravity (g) = 9.8 m/s² (acting downwards)

### Step 2: Use the third equation of motion
We will use the third equation of motion which relates initial velocity, final velocity, acceleration, and displacement (height in this case):
\[ v^2 = u^2 - 2gh \]

### Step 3: Substitute the known values into the equation
Since we are looking for the height (h) at which the stone stops rising (v = 0), we can substitute the known values into the equation:
\[ 0 = (20)^2 - 2 \cdot 9.8 \cdot h \]

### Step 4: Rearrange the equation to solve for height (h)
Rearranging the equation gives us:
\[ 0 = 400 - 19.6h \]
\[ 19.6h = 400 \]
\[ h = \frac{400}{19.6} \]

### Step 5: Calculate the height (h)
Now, we can calculate the height:
\[ h = \frac{400}{19.6} \approx 20.41 \, \text{meters} \]

### Final Answer
The stone will reach a height of approximately **20.41 meters** before it begins to fall.

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