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The height y and the distance x along th...

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:

A

5 m/s

B

12 m/s

C

13 m/s

D

Not obtainable from the data

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To find the velocity with which the projectile is projected, we need to analyze the given equations for height \( y \) and horizontal distance \( x \) in terms of time \( t \). ### Step-by-Step Solution: 1. **Identify the equations**: The equations given are: \[ y = 5t - 8t^2 \quad \text{(1)} \] \[ x = 12t \quad \text{(2)} \] 2. **Find the horizontal component of velocity \( v_x \)**: The horizontal velocity \( v_x \) is the derivative of \( x \) with respect to \( t \): \[ v_x = \frac{dx}{dt} = \frac{d(12t)}{dt} = 12 \, \text{m/s} \] 3. **Find the vertical component of velocity \( v_y \)**: The vertical velocity \( v_y \) is the derivative of \( y \) with respect to \( t \): \[ v_y = \frac{dy}{dt} = \frac{d(5t - 8t^2)}{dt} = 5 - 16t \, \text{(3)} \] 4. **Calculate \( v_y \) at \( t = 0 \)**: To find the initial vertical velocity, substitute \( t = 0 \) into equation (3): \[ v_y(0) = 5 - 16(0) = 5 \, \text{m/s} \] 5. **Calculate the resultant velocity \( v \)**: The resultant velocity \( v \) is given by the Pythagorean theorem: \[ v = \sqrt{v_x^2 + v_y^2} \] Substitute \( v_x = 12 \, \text{m/s} \) and \( v_y(0) = 5 \, \text{m/s} \): \[ v = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \, \text{m/s} \] ### Final Answer: The velocity with which the projectile is projected is \( 13 \, \text{m/s} \). ---
To find the velocity with which the projectile is projected, we need to analyze the given equations for height \( y \) and horizontal distance \( x \) in terms of time \( t \). ### Step-by-Step Solution: 1. **Identify the equations**: The equations given are: \[ y = 5t - 8t^2 \quad \text{(1)} ...
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