An object is thrown upward from the ground with an

hardcoreangelp50

hardcoreangelp50

Answered question

2022-02-01

An object is thrown upward from the ground with an initial velocity of 32fts. What is the maximum height the object obtains using the formula s=16t2+32t, where s = distance above the ground in feet, and t= time in seconds?

Answer & Explanation

Keely House

Keely House

Beginner2022-02-02Added 13 answers

Step 1
The maximum height with respect to time will occur when the derivative of the distance(time) function equals 0
=16t2+32t
dsdt=32t+32
Maximum occurs when
32t+32=0
t=1
When t=1 the object is at a height of
16(1)2+32(1)
=16 (feet)

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