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The equation of projectile is y = 16 x -...

The equation of projectile is `y = 16 x - (5 x^2)/(4)`. Find the horizontal range.

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To find the horizontal range of the projectile given by the equation \( y = 16x - \frac{5x^2}{4} \), we can follow these steps: ### Step 1: Identify the equation of the projectile The given equation is: \[ y = 16x - \frac{5x^2}{4} \] ...
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Knowledge Check

  • The equation of a projectile is y = sqrt(3)x - ((gx^2)/2) the horizontal range is

    A
    `(2g)/ sqrt(3)`
    B
    `(2sqrt(3))/g`
    C
    `g/(2 sqrt(3))`
    D
    `(sqrt(3)g)/2`

  • The equation of projectile is y=16x-(x^(2))/(4) the horizontal range is:-

    A
    16m
    B
    8m
    C
    64m
    D
    12.8m

  • The equation of motion of a projectile is y = 12 x - (3)/(4) x^2 . The horizontal component of velocity is 3 ms^-1 . What is the range of the projectile ?

    A
    18 m
    B
    16 m
    C
    12 m
    D
    21.6 m

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    The range of a projectile launched at an angle of 15^(@) with horizontal is 1.5km. The range of projectile when launched at an angle of 45^(@) to the horizontal is

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