A projectile moves from the ground such that its horizontal displacement is `x=Kt` and vertical displacement is `y=Kt(1-alphat)`, where K and `alpha` are constants and t is time. Find out total time of flight (T) and maximum height attained `(Y_"max")`

Options

A. `T=alpha,Y_"max"=K/(2alpha)`
B. `T=1/alpha,Y_"max"=(2K)/(alpha)`
C. `T=1/alpha,Y_"max"=K/(6alpha)`
D. `T=1/alpha,Y_"max"=K/(4alpha)`

Explanation

To solve the problem, we need to find the total time of flight \( T \) and the maximum height attained \( Y_{\text{max}} \) for the given projectile motion equations:

1. **Horizontal Displacement**:
\[ x = Kt \]

2. **Vertical Displacement**:
\[ y = Kt(1 - \alpha t) \]

### Step 1: Find the Total Time of Flight \( T \)

The total time of flight occurs when the vertical displacement \( y \) returns to zero. Thus, we set the vertical displacement equation to zero:

\[ y = Kt(1 - \alpha t) = 0 \]

This equation can be satisfied in two cases:

1. \( Kt = 0 \) which gives \( t = 0 \) (initial time)
2. \( 1 - \alpha t = 0 \) which gives \( t = \frac{1}{\alpha} \)

Thus, the total time of flight \( T \) is:

\[ T = \frac{1}{\alpha} \]

### Step 2: Find the Maximum Height \( Y_{\text{max}} \)

To find the maximum height, we need to determine when the vertical velocity is zero. The vertical displacement \( y \) is given by:

\[ y = Kt(1 - \alpha t) \]

To find the vertical velocity, we differentiate \( y \) with respect to time \( t \):

\[ \frac{dy}{dt} = K(1 - \alpha t) + Kt(-\alpha) = K(1 - 2\alpha t) \]

Setting the derivative equal to zero to find the time at which the maximum height occurs:

\[ K(1 - 2\alpha t) = 0 \]

This gives:

\[ 1 - 2\alpha t = 0 \implies t = \frac{1}{2\alpha} \]

Now we substitute \( t = \frac{1}{2\alpha} \) back into the vertical displacement equation to find \( Y_{\text{max}} \):

\[ Y_{\text{max}} = K\left(\frac{1}{2\alpha}\right)\left(1 - \alpha\left(\frac{1}{2\alpha}\right)\right) \]

Simplifying this:

\[ Y_{\text{max}} = K\left(\frac{1}{2\alpha}\right)\left(1 - \frac{1}{2}\right) = K\left(\frac{1}{2\alpha}\right)\left(\frac{1}{2}\right) = \frac{K}{4\alpha} \]

### Final Answers

Thus, we have:

- Total Time of Flight \( T = \frac{1}{\alpha} \)
- Maximum Height Attained \( Y_{\text{max}} = \frac{K}{4\alpha} \)