Please answer all steps, make sure the answer is clear and concise, thank you!! Use the filled one answers prior to help answer for the empty and blank steps! Assume the acceleration of the object is a(t) - -9.8 meters per second. (Neglect air resistance.)With what initial velocity must an object be thrown upward (from a height of 3 meters) to reach a maximum height of 360 meters?Step 3We want to find the initial velocity, v(0) of an object thrown upward such that the maximum of the height function t) is 360. We also assume the acceleration due to gravity is a(t) - -9.8 meters per second.First, we must find an equation for the velocity and height functions. Recall that the derivative of the velocity function is acceleration, or v(t) = a(t). This also means that v(t) is the antiderivative of the constant function a(C).Find v(t).v(t) =a(t) dt-9.8 -9.8 t + CThe initial velocity is the velocity at timet- 0. Find the initial velocity by substituting 0 for t in the equation above.V(0) --9.8 Step 2We have found that the velocity of the object thrown upward has the general solution v(t) = -9.8t + C, and the initial velocity is v(0) = C.Substituting vO) for the constant C to find the particular solution for the veloaity gives us the following.vit) = -9.8t + v(0)Now recall that the derivative of the height function is the velocity, or s'T) = v(t). This also means that s(t) is the antiderivative of vit).Find s(t).s(t) =(-9.8t+ v(0)) dt-4.9+ v(0X + C,Step 3We now know that the general solution of the height function is s(t) = -4.9t2 + v(0)t + C,. We also know that the initial height of the object is 3 meters. In other words, s(0) =Use the initial height of the object to solve for the constant C,.s(0) = - 4.9+ v{0)(0} + C,3 =| + 0 + C,C,

Step 2 We have found that the velocity of the object thrown upward has the general solution v(t) = -9.8t + C, and the initial velocity is v(0) = C. Substituting vO) for the constant C to find the particular solution for the velocity gives us the following. v(t) = -9.8t + v(0) Now recall that the derivative of the height function is the velocity, or sTE) = v(t). This also means that s(t) the antiderivative of vit). Find s(t). s(t) = v(0}) dt %3D 4.9 + v{OX + C Step 3 We now know that the general solution of the height function is s(t) - -4.9t2 + v(0)t + C,. We also know that the initial height of the object is 3 meters. In other words, s(0) = Use the initial height of the object to solve for the constant C. s(0) - - 4.9 + v(0)(0) + C, 3 = + 0 + C

Step 2 We have found that the velocity of the object thrown upward has the general solution v(t) = -9.8t + C, and the initial velocity is v(0) = C. Substituting vO) for the constant C to find the particular solution for the velocity gives us the following. v(t) = -9.8t + v(0) Now recall that the derivative of the height function is the velocity, or sTE) = v(t). This also means that s(t) the antiderivative of vit). Find s(t). s(t) = v(0}) dt %3D 4.9 + v{OX + C Step 3 We now know that the general solution of the height function is s(t) - -4.9t2 + v(0)t + C,. We also know that the initial height of the object is 3 meters. In other words, s(0) = Use the initial height of the object to solve for the constant C. s(0) - - 4.9 + v(0)(0) + C, 3 = + 0 + C

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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