A projectile is thrown upward so that its distance above the ground after t seconds is $h = - 16t^2 + 480t$.

After how many seconds does it reach its maximum height?
15 s
22.5 s
7 s
30 s

Question

A projectile is thrown upward so that its distance above the ground after t seconds is $h = - 16t^2 + 480t$.

After how many seconds does it reach its maximum height?
15 s
22.5 s
7 s
30 s

sleepyjess · Answer

@hartnn last one i hope

sleepyjess · Answer

I have done these before but I can't remember how I did it and I can't find my notes on it either

hartnn · Answer

for
$y= ax^2+bx+c$
the maximum occurs at
$-b/2a$

hartnn · Answer

h=−16t^2+480t.
compare this with
h =at^2 +bt+c

find a,b
then use that formula

sleepyjess · Answer

so -480/-32
15

hartnn · Answer

yes :)

sleepyjess · Answer

Thank you so much for all of your help!

hartnn · Answer

welcome ^_^

nincompoop · Answer

there's another way to solve this, but that's probably for another day or semester

nincompoop · Answer

it is to make sense why -b/2a is legitimate

javk · Answer

After 15s
You could also resolve this by imagining its graph
Initially as time increases height also increases
Once maximum height is reached, with any increase in time height decreases
Therefore the graph has zero gradient at maximum height
By differentiation we get:
$$dh/dt=32t+480$$
Where gradient equals zero, therefore dh/dt=0
$$32t+480=0$$
$$t=15$$