107–110. {Use of Tech} Motion with gravity Consider the follow...

Question

107–110. {Use of Tech} Motion with gravity Consider the following descriptions of the vertical motion of an object subject only to the acceleration due to gravity. Begin with the acceleration equation a(t) = v' (t) = -g , where g = 9.8 m/s² .

b. Find the position of the object for all relevant times.

A payload is released at an elevation of 400 m from a hot-air balloon that is rising at a rate of 10 m/s.

Accepted Answer

Welcome back, everyone. A rock is thrown vertically upward from the top of a 250 m cliff with an initial velocity of 15 m per second. Assume that only gravity acts on the rock. G is equal to 9.8 m per second squared. Find the opposition function for all relevant times. For this problem, let's recall that the position function as of T can be identified by integrating the velocity function V of TDT. And also our velocity function can be identified by integrating the acceleration function A of TDT. So we're going to begin with the velocity function. We want to identify the velocity function B of T, and this is going to be equal to the integral of the acceleration function, and we're provided with the acceleration due to gravity. So we want to integrate G of T D T. Now what we have to understand is that. The rock is thrown vertically up, so the initial velocity vector is pointing up, and it's positive, right? We have 15 m per second initially. And acceleration due to gravity is pointing down towards the center of the Earth, right? So this means that G is going to be negative because the direction is going down, right? And therefore, our G of T acceleration due to gravity is -9.8 m per second squared. So what we want to do is simply evaluate the velocity function by integrating -9.8 DT. We can factor out the constant that would be -9.8, integral DT, which is equal to -9.8 T plus a constant of integration C1. And now, the initial condition is that our velocity initially is 15 m per second, meaning velocity at time of zero is 15. So we can set 15 equals negative 9.8 multiplied by 0 plus C1, and from here we can show that the constant of integration C1 is 15. So that the velocity function V of T becomes negative 9.8 T + 15. And now we are ready to evaluate the position functions of T by integrating -9.8 T plus 15. The So, let's split it into two parts. We have -9.8 integral of TDT. Plus 15 integral of DT. And now what we're going to do is simply evaluate the result for each integral. For the first integral, we can apply the power rule. We get -9.8, multiplied by T2 divided by 2, according to the power rule. For the second integral, the integral of the T is going to be equal to T, so we're adding 15 T. And this is an indefinite integral, so we're adding a constant of integration C2, right? We want to identify C2 and we know that the initial position at time of 0 is 250, right? So basically, as of 0 is 250 m, meaning we can set an equation 250 equals -9.8. Divided by 2 witches. -4.9, so we can also simplify rights, so we can say -4.9. Multiplied by 02 + 15 multiplied by 0 + C2. So our constant C2 is equal to 2150. And now we are ready to state the position function as of T equals -4.9. Multiplied by T2. Plus 15 T. Plus our constant of integration which is 250. And that'll be our final answer for this problem. Thank you for watching.

Key Concepts

Acceleration due to Gravity

Velocity and Position Functions

Kinematic Equations

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