The trajectory of a projectile,projected from | vedclass

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(A) The equation of the trajectory is given by $y = x - \frac{x^2}{20}$.To find the maximum height,we need to find the value of $y$ when the slope of

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$5$

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(A) The equation of the trajectory is given by $y = x - \frac{x^2}{20}$.To find the maximum height,we need to find the value of $y$ when the slope of the trajectory is zero,i.e.,$\frac{dy}{dx} = 0$.Differentiating the equation with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(x - \frac{x^2}{20}) = 1 - \frac{2x}{20} = 1 - \frac{x}{10}$.Setting the derivative to zero:$1 - \frac{x}{10} = 0 \Rightarrow x = 10\,m$.Now,substitute $x = 10$ back into the trajectory equation to find the maximum height $y_{\max}$:$y_{\max} = 10 - \frac{(10)^2}{20} = 10 - \frac{100}{20} = 10 - 5 = 5\,m$.

The trajectory of a projectile,projected from the ground,is given by $y = x - \frac{x^2}{20}$,where $x$ and $y$ are measured in meters. The maximum height attained by the projectile will be $...........\,m$.

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