Question

A smart-phone is thrown upwards from the top of a 448-foot building with an initial velocity of 48 feet per second. The height h of the smart-phone after t seconds is given by the quadratic equation h = − 16 t 2 + 48 t + 448 h=-16t2+48t+448. When will the smart-phone hit the ground?

Answer 1

The smart-phone hit the ground when t = 7 s

Explanation:

The height "h" is defined as:

h=16t^2 + 48t + 448

And, when the smart-phone hits the ground, h = 0 ft . Then,

16t^2 + 48t + 448 = 0

And this is a quadratic equation, and we can solve it using the formula for ax^2 + bx + c = 0, which is

x=
`\frac{-b±\sqrt{b^(2)-4ac } }{2a}`

So,

t =
`\frac{-48±\sqrt{48^(2) -4(-16)(448)} }{2(16)}`

t =
`(-48±√(2304+28672) )/(-32)`

And, we have two responses,

t_1 =
`(-48+√(30976) )/(-32)` and t_2 =
`(-48-√(30976) )/(-32)`

t_1 = - 4 s and t_2 = 7 s

As we know, the time is a quantity that cannot have a negative value, so, we take the result 2.