Question

An object is thrown upward from the top of a 160 foot building with an initial velocity of 48 feet per second. The height h of the object after t seconds is given by the quadratic equation h= 16t2+48t+160. When will the object hit the ground?

Answer 1

Answer:

$t=5s$

Explanation:

From free falling objects we know that height is equal to:

$h=h_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

From the exercise we know that

$h_{o}=160ft\\v_{oy}=48ft/s\\g=-32 ft/s^{2}$

Being said that, we can calculate how long would it take to the object to hit the ground knowing that the height at that point is 0

$0=160ft+(48ft/s)t-\frac{1}{2}(32ft/s^2)t^2$

Solving the quadratic formula we got that

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=48\\c=160$

$t=-2s$ or $t=5s$

Since time can not be negative the logical answer is t=5s

So, the object hit the ground at 5s

Answer 2

Answer:

5 s

Explanation:

when the object hits the ground, we're hitting the x-axis, where y = 0.

0 = -16t2 + 48t + 160

Let's divide -16 from both sides...

0 = t2 - 3t - 10

0 = (t - 5)(t + 2)

t - 5 = 0

t = 5 seconds

t + 2 = 0

t = -2

Unless we go back in time 2 seconds, this answer doesn't work.

The object will hit the ground in 5 seconds!