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

Answer:

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

Step-by-step 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 = $\frac{-48±\sqrt{2304+28672} }{-32}$

And, we have two responses,

t_1 = $\frac{-48+\sqrt{30976} }{-32}$    and    t_2 = $\frac{-48-\sqrt{30976} }{-32}$

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.