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

Step-by-step 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)=(-b\pm √(b^2-4ac))/(2a)`
`\mathrm{For\:}\quad a=-16,\:b=45,\:c=204`
`t_(1,\:2)=(-45\pm √(45^2-4\left(-16\right)\cdot \:204))/(2\left(-16\right))`

Apply rule -(-a) = a:

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