A stone thrown upwards from ground level, has its equation of height `h=490t-4.9t^(2)`where`'h'` is in metres and `t` is in seconds respectively. What is the maximum height reached by it ?

Explanation

To find the maximum height reached by the stone thrown upwards, we start with the given equation of height:

\[ h = 490t - 4.9t^2 \]

### Step 1: Differentiate the height function
To find the maximum height, we first need to find the critical points by differentiating the height function with respect to time \( t \).

\[ \frac{dh}{dt} = 490 - 9.8t \]

### Step 2: Set the derivative to zero
Next, we set the derivative equal to zero to find the critical points.

\[ 490 - 9.8t = 0 \]

### Step 3: Solve for \( t \)
Now, we solve for \( t \):

\[ 9.8t = 490 \]
\[ t = \frac{490}{9.8} = 50 \text{ seconds} \]

### Step 4: Verify that it is a maximum
To confirm that this critical point corresponds to a maximum height, we need to check the second derivative.

\[ \frac{d^2h}{dt^2} = -9.8 \]

Since the second derivative is negative, this indicates that the function is concave down at \( t = 50 \) seconds, confirming a maximum.

### Step 5: Calculate the maximum height
Now, we substitute \( t = 50 \) seconds back into the original height equation to find the maximum height.

\[ h = 490(50) - 4.9(50^2) \]

Calculating each term:

\[ h = 24500 - 4.9(2500) \]
\[ h = 24500 - 12250 \]
\[ h = 12250 \text{ meters} \]

### Final Answer
The maximum height reached by the stone is **12250 meters**.

---