Question

An rock is thrown upward from a platform that is 204 feet above ground at 45 feet per second. Use the projectile formula

Answer 1

EXPLANATION

Since we have the equation:

$h=-16t^2+v_0t+h_0$

Where v_0 = 45 feet per second and h_0 = height above ground = 204

Plugging in the terms into the equation:

$h=-16t^2+45t+204$

When the rock hits the ground, the height is of 0 ft, thus we need to isolate the time, as shown as follows:

$0=-16t^2+45t+204$$\mathrm{Switch\:sides}$$-16t^2+45t+204=0$$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$\mathrm{For\:}\quad a=-16,\:b=45,\:c=204$$t_{1,\:2}=\frac{-45\pm \sqrt{45^2-4\left(-16\right)\cdot \:204}}{2\left(-16\right)}$

Apply rule -(-a) = a:

$=\sqrt{45^2+4\cdot \:16\cdot \:204}$$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:16\cdot \:204=13056$$=\sqrt{45^2+13056}$$45^2=2025$$=\sqrt{2025+13056}$$\mathrm{Add\:the\:numbers:}\:2025+13056=15081$$=\sqrt{15081}$$t_{1,\:2}=\frac{-45\pm \sqrt{15081}}{2\left(-16\right)}$$\mathrm{Separate\:the\:solutions}$$t_1=\frac{-45+\sqrt{15081}}{2\left(-16\right)},\:t_2=\frac{-45-\sqrt{15081}}{2\left(-16\right)}$$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$$t=-\frac{-45+\sqrt{15081}}{32},\:t=\frac{45+\sqrt{15081}}{32}$

Expressing as decimals:

$t=-2.43,\text{ t=5.243897}$

Since the solution can not be negative, the only possibility is a positive solution.

Therefore, the rock will hit the ground after 5.24 seconds