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A dynamite blast blows a heavy rock straight up with a launch velocity of 49 m/s (176.4 km/h) (see figure). It reaches a height of s = 49t – 4.9t² meters after t seconds. Smax- 2 v= 0 78.4- D0 1 = ? (u) 481ə

A dynamite blast blows a heavy rock straight up with a launch velocity of 49 m/s (176.4 km/h) (see figure). It reaches a height of s = 49t – 4.9t² meters after t seconds. Smax- 2 v= 0 78.4- D0 1 = ? (u) 481ə

College Algebra
College Algebra
10th Edition
ISBN:9781337282291
Author:Ron Larson
Publisher:Ron Larson
Chapter3: Polynomial Functions
Section3.5: Mathematical Modeling And Variation
Problem 7ECP: The kinetic energy E of an object varies jointly with the object’s mass m and the square of the...
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A dynamite blast blows a heavy rock straight up with a launch velocity of 49 m/s (176.4 km/h) (see figure). It reaches a height of s = 49t – 4.9t² meters after t seconds. Smax v = 0 78.4- D0 1 = ? s = 0! a) How high does the rock go? b) What are the velocity and speed of the rock when it is 78.4 m above the ground on the way up? On the way down? | c) What is the acceleration of the rock at any time t during its flight (after the blast)? d) When does the rock hit the ground again? e) What is the domain of s(tof)? (within the context the problem) f) What is the range of s(t)? (within the context the problem) Formulae: s(t) = 49t – 4.9t² mt, position v(t) = 49 – 9.8t mt/s, speed a(t) = -9.8 mt/s² , acceleration.
Transcribed Image Text:A dynamite blast blows a heavy rock straight up with a launch velocity of 49 m/s (176.4 km/h) (see figure). It reaches a height of s = 49t – 4.9t² meters after t seconds. Smax v = 0 78.4- D0 1 = ? s = 0! a) How high does the rock go? b) What are the velocity and speed of the rock when it is 78.4 m above the ground on the way up? On the way down? | c) What is the acceleration of the rock at any time t during its flight (after the blast)? d) When does the rock hit the ground again? e) What is the domain of s(tof)? (within the context the problem) f) What is the range of s(t)? (within the context the problem) Formulae: s(t) = 49t – 4.9t² mt, position v(t) = 49 – 9.8t mt/s, speed a(t) = -9.8 mt/s² , acceleration.
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