Question

The position time equation for a ball thrown vertically up in air is given by
h = 20t – 10t^2. Find the time at which its height is maximum.​

Answer 1

BODY ATTAINS MAX HEIGHT AT 1 SECOND.

Given:

• Position time equation of a ball is

$h = 20t - 10 {t}^{2}$

To find:

• Time at which the height is maximum?

Calculation:

First of all, let's calculate the first order derivation of height function w.r.t time :

$h = 20t - 10 {t}^{2}$

$\implies \: \dfrac{dh}{dt} = \dfrac{d(20t - 10 {t}^{2}) }{dt}$

$\implies \: \dfrac{dh}{dt} = 20 - 20t$

• Now, for maxima, dh/dt should be equal to zero.

$\implies \: 20 - 20t = 0$

$\implies \: t = 1 \: sec$

• Now, calculating the second order derivative.

$\implies \dfrac{ {d}^{2} h}{d {t}^{2} } = \dfrac{d(20 - 20t)}{dt}$

$\implies \dfrac{ {d}^{2} h}{d {t}^{2} } = - 20$

$\implies \dfrac{ {d}^{2} h}{d {t}^{2} } < 0$

• So, maximum value of h is attained at t = 1 second.

So, body attains maximum height at 1 second.