Question

missile is fired from ground level with an initial velocity of v(0) = 1000, 1400, 368 measured in ft/sec, with the engine shutting off just after launch. If the only external force acting on the missile is the gravitational force, find the trajectory of the missile. (Use g = 32 ft/sec2. Your instructors prefer angle bracket notation < > for vectors.)

Answer 1

Answer:

the trajectory for the missile $\mathbf{r(t) = \langle1000t,1400t, 368t -16t^2 \rangle}$

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:

$r(t) = v_it + \dfrac{1}{2}at^2$

$r(t) = \langle1000,1400,368 \rangle t + \dfrac{1}{2} \langle 0,0, -32\rangle t^2$

$r(t) = \langle1000t,1400t,368t \rangle + \langle 0,0, -16 t^2 \rangle$

$r(t) = \langle1000t,1400t, 368t -16t^2 \rangle$

Thus, the trajectory for the missile $\mathbf{r(t) = \langle1000t,1400t, 368t -16t^2 \rangle}$

To determine the downrange of the missile,

$v(0) = \langle 1000,1400,368 \rangle$

where;

the horizontal vel. of the missle $v_h= \sqrt{1000^2+1400^2}$

= 1720.4651 ft/s

the vertical vel. of the missile is $v_v = 368 \ ft/s$

The time required to reach the ground $t =\dfrac{2 \times v_v}{g}$

$t =\dfrac{2 \times 368}{32}$

t = 23 sec

Finally, the downrange of the missile $R = v_h \times t$

R = 1720.4651  × 23

R = 39570.6973 ft