Find the number of seconds it will take to reach its maximum height. What is this maximum​ height?

2020-08-17 10:45 pm
If an object is projected upward from ground level with an initial velocity of 80 ft per​ sec, then its height in feet after t seconds is given by ​s(t)=−16t2+80t. Find the number of seconds it will take to reach its maximum height. What is this maximum​ height?


The object will take _____second(s) to reach its maximum height.
​(Simplify your​ answer.)

The maximum height reached by the object is _______ feet.
​(Simplify your​ answer.)

回答 (6)

2020-08-18 12:27 am
✔ 最佳答案
Method 1: By completing square

s(t) = -16t² + 80t
s(t) = -16[t² - 5t]
s(t) = -16[t² - 5t + (2.5)²] + 16(2.5)²
s(t) = -16[t - 2.5]² + 100

For all real values of t, -16[t - 2.5]² ≤ 0
Hence, s(t) = -16[t - 2.5]² + 100 ≤ 100
Maximum s(t) = 100 at t = 2.5

Answers:
The object will take _2.5_ second(s) to reach its maximum height.
The maximum height reached by the object is _100_ feet.

====
Method 2: By differentiation

s(t) = -16t² + 80t
s'(t) = -32t + 80 = -32(t - 2.5)
s"(t) = -32

When t = 2.5:
S(2.5) = -16(2.5)² + 80(2.5) = 100
s'(2.5) = 0
s"(2.5) = -32 < 0

Hence, maximum s(t) = 100 at t = 2.5

Answers:
The object will take _2.5_ second(s) to reach its maximum height.
The maximum height reached by the object is _100_ feet.
2020-08-18 12:07 am
Use -b/2a from Quadratic Equation (unless you know differentiation)
t = -80/-32 s = 2.5 s will be max
Then insert into original equation s(t)=−16(80/32)² +80(80/32)
= 100 feet
2020-08-17 11:33 pm
Velocity becomes at the highest point. ds/dt=-32 t+80. t= 80/32 s= 10/4=5/2 s. This is the time. The maximum height is h= 80*5/2-16*5/2*5/2= 200-4*25= 100 m.
2020-08-17 11:29 pm
The function

.......... ​s(t) = −16t²+80t

is a CONCAVE DOWN parabola. That means that its maximum height will occur at the parabola's vertex.

The vertex occurs at t = -80/2(-16) = 2.5 seconds
.................................s(t) = s(2.5) = 100 feet

The object will take 2.5 second(s) to reach its maximum height......ANS
The maximum height reached by the object is 100 feet...................ANS

(​See graph below)
2020-08-17 10:56 pm
There's a couple of ways to do this.  You could do the calculus method and find where the derivative is 0. But that;s not really necessary.  s(t) is a parabola, and the maximum height occurs at the vertex.
The vertex occurs at -b/2a which is -80/(2(-16)) = 80/32 = 5/2 = 2.5 seconds.  Plug 2.5 into s(t) to get the height
You could also have got the answer by finding the roots of the equation.  This one is easy because it factors
into s(t) = t(-16t + 80).  This has roots when t = 0 and when -16t + 80 = 0, which happens when t = 5.
The vertex is at the average of the roots.  so (0 + 5)/2 = 2.5
2020-08-17 10:51 pm
s(t) = -16t^2 + 80t

The vertex of this parabola will provide information about maximum height.

a = -16, b = 80, c =0 for the standard form of a quadratic equation.

t = -b / 2a = -80 / 2(-16) = -80 / -32 = 2.5 seconds for max height.

s(2.5) = -16(2.5^2) + 80(2.5) = 100 feet is max height attained.


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