A body is projected at time `t = 0` from a certain point on a planet's surface with a certain velocity at a certain angle with the planet's surface (assumed horizontal). The horizontal and vertical displacement x and y (in metre) respectively vary with time t in second as, `x= (10sqrt(3)) t and y= 10t - t^2`. The maximum height attained by the body is

Options

A. 75 m
B. 100 m
C. 50 m
D. 25 m

Explanation

To find the maximum height attained by the body, we will analyze the vertical displacement equation given in the problem. The vertical displacement \( y \) is given by:

\[ y = 10t - t^2 \]

### Step 1: Differentiate the vertical displacement equation
To find the time at which the maximum height occurs, we need to differentiate the equation with respect to time \( t \) and set the derivative equal to zero.

\[ \frac{dy}{dt} = \frac{d}{dt}(10t - t^2) \]

### Step 2: Calculate the derivative
Calculating the derivative:

\[ \frac{dy}{dt} = 10 - 2t \]

### Step 3: Set the derivative equal to zero
To find the time at which the maximum height occurs, set the derivative equal to zero:

\[ 10 - 2t = 0 \]

### Step 4: Solve for \( t \)
Now, solve for \( t \):

\[ 2t = 10 \\ t = 5 \text{ seconds} \]

### Step 5: Substitute \( t \) back into the vertical displacement equation
Now that we have the time at which the maximum height occurs, we substitute \( t = 5 \) seconds back into the vertical displacement equation to find the maximum height \( y \):

\[ y = 10(5) - (5)^2 \]

### Step 6: Calculate the maximum height
Calculating the value:

\[ y = 50 - 25 = 25 \text{ meters} \]

### Conclusion
The maximum height attained by the body is:

\[ \boxed{25 \text{ meters}} \]

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