Motion Under Gravity Show that an object thrown from an initial height h0 with an in

Motion Under Gravity Show that an object thrown from an...

$$\begin{array}{c}{\text { Motion Under Gravity Show that an object thrown from an }} \\ {\text { initial height } h_{0} \text { with an initial velocity } v_{0} \text { has a height at time } t} \\ {\text { given by the function }} \\ {h(t)=\frac{1}{2} g t^{2}+v_{0} t+h_{0}}\end{array}$$ where $g$ is the acceleration due to gravity, a constant with value $-32$ ft $/ \mathrm{sec}^{2}$ .

$$\begin{array}{c}{\text { Motion Under Gravity Show that an object thrown from an }} \\ {\text { initial height } h_{0} \text { with an initial velocity } v_{0} \text { has a height at time } t} \\ {\text { given by the function }} \\ {h(t)=\frac{1}{2} g t^{2}+v_{0} t+h_{0}}\end{array}$$
where $g$ is the acceleration due to gravity, a constant with value $-32$ ft $/ \mathrm{sec}^{2}$ .

Key Concepts

Acceleration due to Gravity

The acceleration due to gravity is a constant value that represents the rate at which an object accelerates downward near Earth’s surface. This specific constant is used in the kinematic equations to account for the effect of gravity on the motion of the object.

Initial Conditions

Initial conditions, such as the initial height and initial velocity, are crucial since they provide the specific values needed to determine the constants of integration when deriving equations of motion. They ensure that the general kinematic equation accurately reflects the particular motion situation described.

Uniform Acceleration

In problems involving gravity, the acceleration is constant, meaning the object’s change in velocity over time happens at a steady rate. This concept allows us to use the well-known kinematic equations for uniform acceleration to describe the object's motion.

Integration in Kinematics

The process of integration in kinematics is used to derive equations for velocity and displacement from the known constant acceleration. By integrating the acceleration function with respect to time and applying initial conditions, one obtains the velocity function, and a subsequent integration gives the displacement function.

Transcript

00:01 So we want to show that if we have an object thrown from initial height, h0, with an initial velocity b0, that the height at time t is given by the function h of t is equal to one -half gt squared plus v -0 -t plus h -0, where g is acceleration due to gravity, which is just a constant that has the value negative 32.

00:31 Feet per second squared.

00:36 So first we might want to recall the relationship between some of these things, or maybe first what we should say is that this height function is really just another way of saying displacement.

00:56 So we could think of this as our s of t function, which is how displacement is often written.

01:05 So recall that first, first, the derivative of, or maybe i should write, velocity is equal to the derivative of our displacement.

01:26 So in this case, it will be h of t.

01:32 And if we were to integrate each of these with respect to t, so the derivative and the antiderivative will cancel out, and we'll be left with h of.

01:51 So if we integrate our velocity function, we should get h -t.

01:56 And then we also know that our acceleration function is equal to the derivative of velocity.

02:11 So if we go ahead and integrate this, for the integral of our acceleration and the integral of our velocity will give, so the derivative and integral count so we'll be left with v of t.

02:35 So if we want to find h of t, we're going to have to take our acceleration function and integrate it two times.

02:47 So here we're told that our acceleration is g, which is a constant.

03:01 If i first want to find b of t, because i'll need to find b of t, if i want to find h of t, i want to integrate my acceleration function.

03:12 And i'll just go ahead and plug this in as g, since in the h of t here we also have it as g.

03:19 So this will be the integral of a constant g, b, t.

03:25 And recall that this here is really t to the zero of power.

03:34 We can go ahead and use the scalar property to rewrite this as g integral of t0 dt, and then integrating this will give g to 0 plus 1 divided by 1 plus some constant c, and this here would simplify down to g of t plus c.

04:04 Okay, so now we're told that our initial velocity is v0.

04:16 So that means when our time is zero, we should get v0 out.

04:22 So v of zero is equal to b0, which is equal to.

04:30 So, g of t plus c, and then i plug in zero here.

04:37 And then this here would tell me that v of 0 is equal to c.

04:44 So i have this here...

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