   Chapter 2.3, Problem 4CP ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem

# Checkpoint 4 Worked-out solution available at LarsonAppliedCalculus.comAt time t = 0, a diver jumps from a diving board that is 12 feet high with an initial velocity of 16 feet per second. The diver’s position function is s = −16t2 + 16t + 12.a. When does the diver hit the water?b. What is the diver’s velocity at impact?

(a)

To determine

To calculate: The time when the diver hits the water, when at time t=0 the diver jumps from the diving board which is 12 feet high with an initial velocity of 16 feet per second. The diver's position function is modeled by s=16t2+16t+12.

Explanation

Given Information:

At time t=0 the diver jumps from the diving board which is 12 feet high with an initial velocity of 16 feet per second and the diver's position function is modeled by s=16t2+16t+12.

Formula used:

Zero product law:

If ab=0 then either a=0 or b=0.

Where a, b belongs to set of real numbers.

Calculation:

The position function of the diver is given by

s=16t2+16t+12

To find the time when the diver hits the water, substitute s=0 in the function s=16t2+16t+12.

16t2+16t+12=0

Factor out (4) from the left side of the equation

4(4t24t3)=04t24t3=0

It can be further si

(b)

To determine

To calculate: The velocity at impact when the diver hits the water where the diver's position function is given by s=16t2+16t+12

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