Answer:
[tex]t = 2.57[/tex]
Step-by-step explanation:
Given
[tex]h=-16t^2+12t+75[/tex]
Required
Determine the time to hit the ground
The ground is at point 0.
So, to solve this; we simply set [tex]h =0[/tex], then calculate the value of t
[tex]0=-16t^2+12t+75[/tex]
Multiply through by -1
[tex]0=16t^2-12t-75[/tex]
Reorder
[tex]16t^2-12t-75 = 0[/tex]
Solve using quadratic formula:
[tex]t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]
Where
[tex]a = 16; b = -12; c = -75[/tex]
[tex]t = \frac{-(-12)\±\sqrt{(-12)^2 - 4*16*(-75)}}{2*16}[/tex]
[tex]t = \frac{12\±\sqrt{144 +4800}}{32}[/tex]
[tex]t = \frac{12\±\sqrt{4944}}{32}[/tex]
[tex]t = \frac{12\±70.31}{32}[/tex]
Split:
[tex]t = \frac{12+70.31}{32}[/tex] or [tex]t = \frac{12-70.31}{32}[/tex]
[tex]t = \frac{82.31}{32}[/tex] or [tex]t = \frac{-58.31}{32}[/tex]
But time (t) can't be negative;
So, we make use of only
[tex]t = \frac{82.31}{32}[/tex]
[tex]t = 2.57[/tex]
Hence, the time to hit the ground is 2.57 seconds
