Substitute s, = 2 into the function f(t) = -4.9t2 + vot + Sp. f(t) = -4.9t2 + vot + 2v When the object is at the given maximum height we have -4.9t2 + vat = 388 V 388 Step 3 Now, differentiate v f(t) with respect to t , to obtain the velocity v(t) as a function of time t, v(t) = t + Vo: When the object reaches the topmost point, it comes to rest (before starting to fall down). Hence, it's velocity, v(t), at that point of time is m/sec. Substitute this value for v(t) in the expression for v(t) and simplify to obtaint in terms of vo- Vo Submit Skip (you cannot come back)

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Substitute s, = 2 into the function f(t) = -4.9t² + vot + so:
f(t) = -4.9t2 + vot +
When the object is at the given maximum height we have
-4.9t? + vt = 388 V
388.
Step 3
Now, differentiate
f(t) with respect to t
, to obtain the velocity v(t) as a function of time t,
v(t) =
t + Vo
When the object reaches the topmost point, it comes to rest (before starting to fall down). Hence, it's velocity, v(t), at that point of time is
m/sec.
Substitute this value for v(t) in the expression for v(t) and simplify to obtain t in terms of v..
Submit || Skip (you cannot come back)
Transcribed Image Text:Substitute s, = 2 into the function f(t) = -4.9t² + vot + so: f(t) = -4.9t2 + vot + When the object is at the given maximum height we have -4.9t? + vt = 388 V 388. Step 3 Now, differentiate f(t) with respect to t , to obtain the velocity v(t) as a function of time t, v(t) = t + Vo When the object reaches the topmost point, it comes to rest (before starting to fall down). Hence, it's velocity, v(t), at that point of time is m/sec. Substitute this value for v(t) in the expression for v(t) and simplify to obtain t in terms of v.. Submit || Skip (you cannot come back)
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