Question

The question asks 'What is the maximum height attained by the pebble?' Do I plug in 3 in place of t ?

http://i.imgur.com/sWZyK7G.jpg is the full question, but I posted more information below.

Answer 1

\[h(t) = -16t ^{2} + v _{0} t +h _{0}\]

Is the formula, \[h _{0} =48\]
and
\[v _{0} =64\].

The question asks 'What is the maximum height attained by the pebble?'

Our roots from the quadratic equation are 1 and 3. How do I solve for the answer?

Answer 2

using the equation, we get the roots when h(t) = 0....

Answer 3

if that is the case...
h(t) = 0 = -16t^2 + 64t + 48 = -t^2 + 4t + 3 = t^2 - 4t - 3
and solving the equation yields 1 and 3 as the roots...

Answer 4

and to answer, the original problem... we apply time rates or calculus...

Answer 5

if we differentiate h(t) with t...\[h(t) = -16t ^{2}+v_{0}t+h _{0}\]

Answer 6

\[\frac{ dh(t) }{ dt }=-32t+64\]

Answer 7

the 1st derivative is actually our equation of velocity where 64 fps is just the initial velocity... at maximum height, our velocity is zero (0)...
\[\frac{ d(h) }{ dt }=0=-32t+64\]

Answer 8

we can now compute the time it takes the pebble reach the maximum height...

Answer 9

the time t is 2 seconds...

Answer 10

if we plug 2 in place of t in our original quadratic equation, maximum height is 112 feet...

Answer 11

\[h _{\max}(2) = -16(2)^{2}+64(2)+48=112 feet\]

Answer 12

How did you get t ? You said it was 2 seconds but how did you reach that with the equation regarding the 1st derivative?

Answer 13

The problem is a time rate problem in calculus, if the given equation refers to a distance such as h, the 1st derivative refers to velocity of pebble going up (as thrown upward) and down (for the return)... and in calculus equating 1st derivative to zero(0) refers either to a maximum or minimum value... that is the reason why we equate 1st derivative to zero...

Answer 14

Ah okay, I see now. Thank you.

Answer 15

setting dh/dt = 0, allows us to determine the time t it takes for the pebble at maximum height since before the pebble returns down, its velocity at exactly time t will come to stop and then increase again downward due to the pull of gravity...

Answer 16

time t at max height again is 2 seconds as per our calculation....

Answer 17

Oh wow, yeah I'm sorry I missed it. When you set dh/dt to 0 you just solve for t. Add positive 32t to both sides and divide both sides by 32.

My last question is how did you get to -32? 64 is the initial velocity but is -32 just our 2a from the quadratic equation?

Answer 18

applying power formula in calculus....

Answer 19

\[d(x ^{n}) = nx ^{n-1} dx\]

Answer 20

\[d(-16t ^{2})=2*(-16)t ^{2-1}=-32t\]

Answer 21

our equation for height h...
\[h(t) =-16t ^{2}+64t+48\]

Answer 22

our equation for velocity v...
\[\frac{ dh }{ dt }=v=-32t+64\]

Answer 23

Ah okay, that makes things alot more understandable.

Answer 24

to find the height at any time t, we use h(t) equation, and to find velocity at any time t we use v equation derived....

Answer 25

Got it. Thank you very much for your time.

Answer 26

no problem... :)