Answer:
[tex]t=5s[/tex]
Explanation:
From free falling objects we know that height is equal to:
[tex]h=h_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]
From the exercise we know that
[tex]h_{o}=160ft\\v_{oy}=48ft/s\\g=-32 ft/s^{2}[/tex]
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
[tex]0=160ft+(48ft/s)t-\frac{1}{2}(32ft/s^2)t^2[/tex]
Solving the quadratic formula we got that
[tex]t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}[/tex]
[tex]a=-16\\b=48\\c=160[/tex]
[tex]t=-2s[/tex] or [tex]t=5s[/tex]
Since time can not be negative the logical answer is t=5s
So, the object hit the ground at 5s
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!
