Answer:
the trajectory for the missile [tex]\mathbf{r(t) = \langle1000t,1400t, 368t -16t^2 \rangle}[/tex]
the downrange of the missile R = 39570.6973 ft
Explanation:
Suppose the initial velocity v(0) = <1000, 1400, 368 > ft/sec
The single force which is acting over the missle is the gravitational force.
a (t) = <0, 0, -32> ft/sec²
We are to determine the trajectory and the downrange for the missle
Consider the motion of a given object r(t) to be:
[tex]r(t) = v_it + \dfrac{1}{2}at^2[/tex]
[tex]r(t) = \langle1000,1400,368 \rangle t + \dfrac{1}{2} \langle 0,0, -32\rangle t^2[/tex]
[tex]r(t) = \langle1000t,1400t,368t \rangle + \langle 0,0, -16 t^2 \rangle[/tex]
[tex]r(t) = \langle1000t,1400t, 368t -16t^2 \rangle[/tex]
Thus, the trajectory for the missile [tex]\mathbf{r(t) = \langle1000t,1400t, 368t -16t^2 \rangle}[/tex]
To determine the downrange of the missile,
[tex]v(0) = \langle 1000,1400,368 \rangle[/tex]
where;
the horizontal vel. of the missle [tex]v_h= \sqrt{1000^2+1400^2}[/tex]
= 1720.4651 ft/s
the vertical vel. of the missile is [tex]v_v = 368 \ ft/s[/tex]
The time required to reach the ground [tex]t =\dfrac{2 \times v_v}{g}[/tex]
[tex]t =\dfrac{2 \times 368}{32}[/tex]
t = 23 sec
Finally, the downrange of the missile [tex]R = v_h \times t[/tex]
R = 1720.4651 × 23
R = 39570.6973 ft
The trajectory of the missile as a function of time is obtained as
[tex]r(t)= <100t, 1400t, 368t-16t^2>[/tex].
Given that the initial velocity is;
[tex]v(0) = <1000, 1400, 368>\,ft/sec[/tex]
The acceleration of the missile is provided by the gravitational force.
[tex]a = g = <0, 0, -32>\,ft/sec^2[/tex]
The trajectory of the object at time 't' is given by;
[tex]r(t)=v(0)\,t+\frac{1}{2} at^2= <100, 1400, 368>t\,+\frac{1}{2} <0, 0, -32>t^2[/tex]
[tex]\implies r(t)= <100t, 1400t, 368t-16t^2>[/tex]
Learn more about projectile motion here:
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