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_max) by the projectile given the equations for horizontal and vertical displacements.

### Step 1: Understand the equations
The horizontal displacement is given by:
\[ x = Kt \]
The vertical displacement is given by:
\[ y = Kt(1 - \alpha t) \]
where \( K \) and \( \alpha \) are constants, and \( t \) is time.

### Step 2: Find the time of flight (T)
The time of flight is the total time the projectile is in the air. At the time of flight, the vertical displacement \( y \) returns to 0. We set the vertical displacement equation to zero:
\[ 0 = Kt(1 - \alpha t) \]

This equation can be factored as:
\[ Kt = 0 \quad \text{or} \quad 1 - \alpha t = 0 \]

From \( Kt = 0 \), we get:
\[ t = 0 \] (initial position)

From \( 1 - \alpha t = 0 \), we can solve for \( t \):
\[ \alpha t = 1 \]
\[ t = \frac{1}{\alpha} \]

Thus, the total time of flight \( T \) is:
\[ T = \frac{1}{\alpha} \]

### Step 3: Find the maximum height (Y_max)
To find the maximum height, we need to determine the time at which the vertical component of the velocity is zero. The vertical displacement equation is:
\[ y = Kt(1 - \alpha t) \]

To find the velocity, we differentiate \( y \) with respect to \( t \):
\[ \frac{dy}{dt} = K(1 - \alpha t) + Kt(-\alpha) \]
\[ \frac{dy}{dt} = K(1 - 2\alpha t) \]

Setting the vertical velocity to zero to find the time at maximum height:
\[ 0 = K(1 - 2\alpha t) \]

From this, we get:
\[ 1 - 2\alpha t = 0 \]
\[ 2\alpha t = 1 \]
\[ t = \frac{1}{2\alpha} \]

Now we substitute this time back into the vertical displacement equation to find the maximum height:
\[ Y_{max} = K\left(\frac{1}{2\alpha}\right)\left(1 - \alpha\left(\frac{1}{2\alpha}\right)\right) \]
\[ Y_{max} = K\left(\frac{1}{2\alpha}\right)\left(1 - \frac{1}{2}\right) \]
\[ Y_{max} = K\left(\frac{1}{2\alpha}\right)\left(\frac{1}{2}\right) \]
\[ Y_{max} = \frac{K}{4\alpha} \]

### Final Answers
- Total time of flight \( T = \frac{1}{\alpha} \)
- Maximum height attained \( Y_{max} = \frac{K}{4\alpha} \)