calculus question...I REALLY NEED YOUR HELP!!?

Question

I'm kinda stuck...wait I take that back I'm really stuck!

The question is

A skyrocket is fired vertically upwards. Its height after [t] is
s[t] = 60t - 5t^2

[s] is measured in metres and [t] is measured in seconds.

Find the maximum height of the skyrocket.

Answer 1

as u have:
s[t]=60t-5t^2
differentiate it w.r.t to t
d[s(t)]/dt= 60-10t

for maximum height

d[s(t)]/dt= 0

60-10t=0

-10t=-60

t=6sec

put t=6sec in your given equation

s(6) = 60*6 - 5*(6)^2
= 360-180
=180m
this will be your maximum height of the skyrocket.

Answer 2

The equation: s(t) = 60t -5t^2 means the distance the skyrocket had traveled.
If you differentiate the distance, you'll get the speed or velocity.
Therefore velocity is:
s(t)' = 60 - 10t

You have to remember that when the rocket reached its maximum height, velocity is equal to zero. Therefore:
0 = 60 - 10t
-60 = -10t
t = 6

Since the time it takes to reach its max. height is 6 seconds, plug t = 6 into the original formula (the distance)

s(t) = 60t - 5t^2
= 60(6) - 5(6)^2
= 180 m

Answer 3

Okay. You have displacement or position as a function of time. That would be s[t] = 60t - 5t^2

Now, the velocity as a function of time can be derived from the s[t] function by differentiating it. that is, s'[t] is the velocity function.

I think a good deal of visualization can go a long way. At the peak of the rocket's trajectory,or at the maximum height, the velocity of the rocket must be ZERO. So just equate s'[t]=0 and then solve:
s'[t]= 60 - 10t = 0
Therefore, t= 6 seconds. So the maximum height is s[6]= 180 m

Remember: Visualization.

Answer 4

Ok to find maximum height, u can do many methods
1) Equate it to zero
60t - 5t^2 = 0
12t - t^2 = 0
t(12-t) = 0
t=0sec or t=12sec
Therefore the maximum point would be at halfway when t=6sec
when t=6sec,
s[6] = 60(6) - 5(6)^2
= (your answer)


2) An alternative approach would be differentiating your equation to find maximum point.
s[t] = 60t - 5t^2
ds/dt = 60 - 10t
60 - 10t = 0
t=6sec
s[6] = 60(6) - 5(6)^2
= (your answer)

Answer 5

try the derivation...
s[t] = 60t - 5t^2
ds/dt=60 - 10t
for the maximum ds/dt should be 0
60-10t = 0
t=60/10
t=6

then
s[t]=60t-5t^2
s[6]=60x6 - 5x6^2
s[6]=240 - 180
s[6]=60

the maximum heigth is 60 metres.

Answer 6

s(t) = 60 t - 5 t ²
s `(t) = 60 - 10 t = 0 for maximum height
t = 6
s (max) = 360 - 5 x 36
s (max) = 360 - 180
s (max) = 180
Maximum height = 180 m

Answer 7

find the deriative of the equation and set it equal to 0
s'(t) = 60 - 10t

0 = 60 - 10t
-60 = -10t
t = 6

plug 6 back in the origional equation
s(6) = 60(6) - 5(6)^2
s(6) = 180 m