3.4 Motion in space  (Page 4/17)

Newton’s second law also tells us that $F=m\text{a},$ where a represents the acceleration vector of the object. This force must be equal to the force of gravity at all times, so we therefore know that

Now we use the fact that the acceleration vector is the first derivative of the velocity vector. Therefore, we can rewrite the last equation in the form

By taking the antiderivative of each side of this equation we obtain

for some constant vector ${\text{C}}_1.$ To determine the value of this vector, we can use the velocity of the object at a fixed time, say at time $t=0.$ We call this velocity the initial velocity : $\text{v}\left(0\right)={\text{v}}_0.$ Therefore, $\text{v}\left(0\right)=\text{−}g\left(0\right)\text{j}+{\text{C}}_1={\text{v}}_0$ and ${\text{C}}_1={\text{v}}_0.$ This gives the velocity vector as $\text{v}\left(t\right)=\text{−}gt\text{j}+{\text{v}}_0.$

Next we use the fact that velocity $\text{v}\left(t\right)$ is the derivative of position $\text{s}\left(t\right).$ This gives the equation

Taking the antiderivative of both sides of this equation leads to

with another unknown constant vector ${\text{C}}_2.$ To determine the value of ${\text{C}}_2,$ we can use the position of the object at a given time, say at time $t=0.$ We call this position the initial position : $\text{s}\left(0\right)={\text{s}}_0.$ Therefore, $\text{s}\left(0\right)=\text{−}\left(1\text{/}2\right)g{\left(0\right)}^2\text{j}+{\text{v}}_0\left(0\right)+{\text{C}}_2={\text{s}}_0$ and ${\text{C}}_2={\text{s}}_0.$ This gives the position of the object at any time as

Let’s take a closer look at the initial velocity and initial position. In particular, suppose the object is thrown upward from the origin at an angle $\theta$ to the horizontal, with initial speed $v_0.$ How can we modify the previous result to reflect this scenario? First, we can assume it is thrown from the origin. If not, then we can move the origin to the point from where it is thrown. Therefore, ${\text{s}}_0=0,$ as shown in the following figure.

We can rewrite the initial velocity vector in the form ${\text{v}}_0=v_0\text{cos}\theta \text{i}+v_0\text{sin}\theta \text{j}.$ Then the equation for the position function $\text{s}\left(t\right)$ becomes

The coefficient of i represents the horizontal component of $\text{s}\left(t\right)$ and is the horizontal distance of the object from the origin at time t. The maximum value of the horizontal distance (measured at the same initial and final altitude) is called the range R . The coefficient of j represents the vertical component of $\text{s}\left(t\right)$ and is the altitude of the object at time t. The maximum value of the vertical distance is the height H .