Question #beffb

Question

1292 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-21T20:26:45

a) $t = 2 \ s$
b) $t = 0 \ s$
c) $v=0 \ ms^(-1)$

We have height (or displacement) at time $t$ , where $h=0$ is at ground level, and +ve is measured upwards, is given by:

$h(t) = -4.9t^2+9.8t$

Differentiating wrt $t$ will give us the speed:

$doth(t) = -9.8t+9.8$

a) When is the rocket travelling at 9.8 m/s down?

$doth(t) = -9.8$
$:. -9.8t+9.8 = -9.8$
$:. -9.8t = -19.6$
$:. t = 2 \ s$

b) When is the rocket travelling at 9.8m/s up?

$doth(t) = 9.8$
$:. -9.8t+9.8 = 9.8$
$:. -9.8t = 0$
$:. t = 0 \ s$

c) How fast is the rocket travelling after 1 second?

When $t=1$ :

$doth(t) = -9.8+9.8$
$" "= 0 \ ms^(-1)$