Answer:
The object will hit the ground after 9 seconds.
Step-by-step explanation:
We are given the function:
[tex]\displaystyle h(t) = -16t^2 + v_0t + h_0[/tex]
Where h is the height of the object after time t, v₀ is the initial velocity, and h₀ is the initial height.
An object is thrown upwards from a height of 720 feet and an initial velocity of 64ft/s. We can to determine how long it will take for the object to reach the ground.
Hence, h₀ = 720 and v₀ = 64. Substitute:
[tex]\displaystyle h(t) = -16t^2+64t+720[/tex]
When the object reaches the ground, its height h above the ground will be zero. In other words:
[tex]\displaystyle 0 = -16t^2 + 64t + 720[/tex]
Solve for t:
[tex]\displaystyle \begin{aligned} -16t^2 + 64t + 720 &= 0 \\ \\ t^2 - 4t-45 & = 0 \\ \\ (t-9)(t+5) & = 0 \\ \\ t &= -5 \text{ or } t = 9\end{aligned}[/tex]
Since time cannot be negative, we can ignore the first solution.
In conclusion, the object will hit the ground after 9 seconds.
