Question

Need help finding the height and when will object reach the ground?

Answer 1

Height:

`h=-16t^2+73t+18`

As we are asked the time when h = 91ft, then we have to set the equation to 91:

`91=-16t^2+73t+18`

Now, we can solve for t by using the General Quadratic Equation:

`x_(1,2)=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

in which the variables represent the coefficients of a quadratic equation in the form:

`ax^2+bx+c=0`

Therefore, we have to set our equation to 0:

`0=-16t^2+73t+18-91`

Simplifying:

`0=-16t^2+73t-73`

Thus, in our case:

• a = -16

,

• b = 73

,

• c = 73

Replacing these values in the formula:

`t_(1,2)=\frac{-73\pm\sqrt[]{73^2-4\cdot(-16)\cdot(-73)}}{2\cdot(-16)}`

Simplifying:

`t_(1,2)=\frac{-73\pm\sqrt[]{657}}{-32}`
`t_1=\frac{-73+3\sqrt[]{73}}{-32}\approx3.08s`
`t_2=\frac{-73-3\sqrt[]{73}}{-32}\approx1.48s`

As the object is thrown at 73ft/s, at 3 seconds it would be more or less three times higher than 73ft.

Answer: 1.48s