# A target plane flies at 2100 meters. An antiaircraft gun fires a projectile with an initial upward velocity of 200 m/sec. Will the projectile ever reach the altitude of the target plane?

NO, because Maximum height is only
$s = 2 , 040.82 \text{ }$meters

#### Explanation:

From the given : height of the plane$= 2100$ meters

From Physics, $s = {v}_{o} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$ and ${v}_{o} = 200 \text{ } \frac{m}{\sec}$
The $g = - 9.8 \text{ }$$\frac{m}{\sec} ^ 2$(negative because accelaration due to gravity is directed downward)

$s = {v}_{o} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$

$s = 200 \cdot t + \frac{1}{2} \cdot - 9.8 \cdot {t}^{2}$

To determine the maximum, we convert this parabolic equation to its Vertex Form by completing the square

$s = - \frac{9.8}{2} \cdot {t}^{2} + 200 t$

$s = - \frac{9.8}{2} \left({t}^{2} - \frac{400}{9.8} \cdot t\right)$

$s = - \frac{9.8}{2} \left({t}^{2} - \frac{400}{9.8} \cdot t + {\left(\frac{200}{9.8}\right)}^{2} - {\left(\frac{200}{9.8}\right)}^{2}\right)$

$s = - \frac{9.8}{2} {\left(t - \frac{200}{9.8}\right)}^{2} + \frac{9.8}{2} \cdot {\left(\frac{200}{9.8}\right)}^{2}$

simplify first

$s = - \frac{9.8}{2} {\left(t - \frac{200}{9.8}\right)}^{2} + \frac{20000}{9.8}$
$s - \frac{20000}{9.8} = - \frac{9.8}{2} {\left(t - \frac{200}{9.8}\right)}^{2}$

${\left(t - \frac{200}{9.8}\right)}^{2} = - \frac{2}{9.8} \left(s - \frac{20000}{9.8}\right)$

This is the vertex form ${\left(x - h\right)}^{2} = - 4 p \left(y - k\right)$

Maximum $s = \frac{20000}{9.8} = 2040.82 \text{ }$meters

This is not enough to hit the plane at 2100 meters

God bless.... I hope the explanation is useful.