The horizontal and vertical displacement of a projecticle are given by `x=12t` and `y=16t-5t^(2)`, all the quantities being measured in S.I. system. The maximum height of the projectile is `(g=10ms^(-2))`

Options

A. 25.6 m
B. 12.8 m
C. 64 m
D. 6.4 m

Answer

To find the maximum height of the projectile given the equations of horizontal and vertical displacement, we can follow these steps:

### Step 1: Understand the equations of motion
The horizontal displacement is given by:
\[ x = 12t \]
The vertical displacement is given by:
\[ y = 16t - 5t^2 \]

### Step 2: Determine the time at which the projectile reaches maximum height
At maximum height, the vertical component of the velocity becomes zero. To find this, we need to differentiate the vertical displacement equation with respect to time \( t \):
\[ \frac{dy}{dt} = 16 - 10t \]

### Step 3: Set the vertical velocity to zero
To find the time at which the projectile reaches its maximum height, set the derivative equal to zero:
\[ 16 - 10t = 0 \]

### Step 4: Solve for \( t \)
Rearranging the equation gives:
\[ 10t = 16 \]
\[ t = \frac{16}{10} = \frac{8}{5} \text{ seconds} \]

### Step 5: Substitute \( t \) back into the vertical displacement equation
Now, substitute \( t = \frac{8}{5} \) into the vertical displacement equation to find the maximum height:
\[ y = 16\left(\frac{8}{5}\right) - 5\left(\frac{8}{5}\right)^2 \]

### Step 6: Calculate the maximum height
Calculating the first term:
\[ 16 \times \frac{8}{5} = \frac{128}{5} \]

Calculating the second term:
\[ 5 \times \left(\frac{8}{5}\right)^2 = 5 \times \frac{64}{25} = \frac{320}{25} = \frac{64}{5} \]

Now substitute these back into the equation:
\[ y = \frac{128}{5} - \frac{64}{5} = \frac{128 - 64}{5} = \frac{64}{5} \]

### Step 7: Convert to decimal
Now, convert \( \frac{64}{5} \) to decimal:
\[ \frac{64}{5} = 12.8 \text{ meters} \]

### Conclusion
The maximum height of the projectile is \( 12.8 \) meters.

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