The height y and the distance x along the horizontal plane of a projectile on a certain planet [with no surrounding atmosphere] are given by `y=[5t-8t^2]` metre and x = 12t metre where t is the time in second. The velocity with which the projectile is projected is:

Options

A. 5 m/s
B. 12 m/s
C. 13 m/s
D. Not obtainable from the data

Explanation

To find the velocity with which the projectile is projected, we will follow these steps:

### Step 1: Identify the equations of motion
The height \( y \) and horizontal distance \( x \) of the projectile are given by:
\[ y = 5t - 8t^2 \]
\[ x = 12t \]

### Step 2: Differentiate the horizontal distance to find the horizontal component of velocity
To find the horizontal component of the velocity \( u_x \), we differentiate \( x \) with respect to time \( t \):
\[ u_x = \frac{dx}{dt} = \frac{d(12t)}{dt} = 12 \, \text{m/s} \]

### Step 3: Differentiate the height to find the vertical component of velocity
Next, we differentiate \( y \) with respect to time \( t \) to find the vertical component of velocity \( u_y \):
\[ u_y = \frac{dy}{dt} = \frac{d(5t - 8t^2)}{dt} = 5 - 16t \]

### Step 4: Calculate the vertical component of velocity at \( t = 0 \)
Now, we need to find \( u_y \) at \( t = 0 \):
\[ u_y = 5 - 16(0) = 5 \, \text{m/s} \]

### Step 5: Calculate the magnitude of the initial velocity
The initial velocity \( u \) can be calculated using the components \( u_x \) and \( u_y \):
\[ u = \sqrt{u_x^2 + u_y^2} = \sqrt{(12)^2 + (5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \, \text{m/s} \]

### Final Answer
The velocity with which the projectile is projected is:
\[ \boxed{13 \, \text{m/s}} \]

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