To solve this question, follow the steps below.
Step 01: Substitute V0 by 112 feet per second.
[tex]s=-16t^2+112t[/tex]
Step 02: Find t when s = 96.
To do it, first, substitute s by 96:
[tex]96=-16t^2+112t[/tex]
Now, subtract 96 from both sides of the equation:
[tex]\begin{gathered} 96-96=-16t^2+112t-96 \\ 0=-16t^2+112t-96 \\ -16t^2+112t-96=0 \end{gathered}[/tex]
Use the Quadratic formula to find t.
For a equation ax² + bx + c = 0, x is:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
In this question,
a = -16
b = 112
c = -96
Substituting in the equation to find t:
[tex]\begin{gathered} t=\frac{-112\pm\sqrt{112^2-4*(-16)*(-96)}}{2*(-16)} \\ t=\frac{-112\pm\sqrt{12544-6144}}{-32} \\ t=\frac{-112\operatorname{\pm}\sqrt{6400}}{-32} \\ t=\frac{-112\pm80}{-32} \\ t_1=\frac{-112-80}{-32}=\frac{-192}{-32}=6 \\ t_2=\frac{-112+80}{-32}=\frac{-32}{-32}=1 \end{gathered}[/tex]
Time to reach 96 feet = 1 or 6 seconds.
Step 03: Find the time when the projectile will return to the ground.
When the projectile return to the ground, s = 0.
Then, substitute s by 0 and find x using the same equation from step 02.
[tex]\begin{gathered} 0=-16t^2+112t \\ -16t^2+112t=0 \end{gathered}[/tex]
Then,
a = -16
b = 112
c = 0
Substituting in the quadratic formula:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ t=\frac{-112\pm\sqrt{112^2-4*(-16)*0}}{2*(-16)} \\ t=\frac{-112\pm\sqrt{12544-0}}{-32} \\ t=\frac{-112\pm\sqrt{12544}}{-32} \\ t=\frac{-112\pm112}{-32} \\ t_1=\frac{-112-112}{-32}=\frac{-224}{-32}=7 \\ t_2=\frac{-112+112}{-32}=0 \end{gathered}[/tex]
The projectile will reach the ground when t = 0 and when t = 7. t = 0 is the time when the projectile is launched and t = 7 is when the projectile returns to the ground.
(b) So, the projectile returns to the ground when t = 7 seconds.
Answer:
(a) The projectile reached 96 feet when t = 1 or t = 6 seconds.
(b) The projectile returns to the ground when t = 7 seconds.
