Answer to Question #167020 in Calculus for Samir khan

Question

An object is projected directly up so that its height in metres at time t seconds can be modelled by

h(t) = -0.5t2 + 9t + 9.1

a. From what height was the object initially projected?

b. What was the initial velocity?

c. Find the velocity and height when t = 9s.

d. When does the object return to its initial height?

Answer 1

The height of the object in meters at time $t$ seconds is given by $h(t)=-0.5t^2+9t+9.1$

(a) At the initial height $t=0.$

$\therefore h(0)=9.1$

Therefore the object initially projected from the height $9.1$ meters.

(b) Velocity $=$ Rate of change of height $=$ $\frac{dh}{dt}$ .

$\therefore \frac{dh}{dt}=[(-0.5)×2t]+9=(-t+9)$

For the initial Velocity $t=0.$

$\therefore [\frac{dh}{dt}] _{t=0}=9$

Therefore the initial Velocity of the object is $9$ $m/sec.$

(c) At $t=9s$ , $h(9)=[(-0.5)×9^2]+(9×9)+9.1$

$=(-40.5)+81+9.1$

$=49.6$

and $[\frac{dh}{dt}]_{t=9}= (-9+9)=0$

Therefore at $t=9s$ , the height of the object is $49.6$ meters and Velocity is $0.$

(d) The initial height of the object is $9.1$ meters.

According to the problem, $9.1=(-0.5)t^2+9t+9.1$

$\implies 0.5t^2=9t$

$\implies t^2-18t=0$

$\implies t=0$ and $t=18$

But $t=0$ is the initial position of the object.

So the object return to its initial height at $t=18$ $sec.$