# Height attained by particle in vertical direction changing with time y= 5t-5t^2 then maximum height attained by particle is?

Mar 12, 2017

The maximum height is $= 1.25 m$

#### Explanation:

We determine the first and second derivatives

$y = 5 t - 5 {t}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 5 - 10 t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 0$ when $5 - 10 t = 0$

That is,

$5 = 10 t$

$t = \frac{1}{2}$

$\frac{{d}^{2} y}{\mathrm{dt}} ^ 2 = - 10$

As, $\frac{{d}^{2} y}{\mathrm{dt}} ^ 2 < 0$, this is a maximum

$y \left(2\right) = 5 \cdot \frac{1}{2} - 5 \cdot \frac{1}{4} = \frac{5}{4} = 1.25 m$

graph{5x-5x^2 [-3.442, 3.486, -0.016, 3.448]}