The height y and horizontal distance x covered by a projectile in a time t seconds are given by the equations y = `8t -5t^(2)` and x = 6t. If x and y are measured in metres, the velocity of projection is

Options

A. `8 ms^(-1)`
B. `6 ms^(-1)`
C. `14 ms^(-1)`
D. `10 ms^(-1)`

Answer

To find the velocity of projection of a projectile given the equations for height \( y \) and horizontal distance \( x \), we can follow these steps:

### Step 1: Understand the equations
The equations provided are:
- \( y = 8t - 5t^2 \) (for height)
- \( x = 6t \) (for horizontal distance)

### Step 2: Determine the vertical component of the velocity
To find the vertical component of the velocity (\( v_y \)), we need to differentiate the height equation \( y \) with respect to time \( t \):
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(8t - 5t^2) \]
Calculating the derivative:
\[ v_y = 8 - 10t \]

### Step 3: Calculate \( v_y \) at \( t = 0 \)
Substituting \( t = 0 \) into the equation for \( v_y \):
\[ v_y(0) = 8 - 10(0) = 8 \, \text{m/s} \]

### Step 4: Determine the horizontal component of the velocity
To find the horizontal component of the velocity (\( v_x \)), we differentiate the horizontal distance equation \( x \) with respect to time \( t \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(6t) \]
Calculating the derivative:
\[ v_x = 6 \, \text{m/s} \]

### Step 5: Calculate the resultant velocity of projection
The resultant velocity (\( v \)) can be found using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Substituting the values we found:
\[ v = \sqrt{(6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{m/s} \]

### Final Answer
The velocity of projection is \( 10 \, \text{m/s} \).

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