Use the position equation $s = - 20 t ^ { 2 } + v _ { 0 } t + s _ { 0 }$ , where $S$ represents the height of an object (in feet), $v _ { 0 }$ represents the initial velocity of the object (in feet per second), $s _ { 0 }$ represents the Initial height of the object (in feet), and $t$ represents the time (in seconds).
A projectile is fired straight upward from ground level $\left( s _ { 0 } = 0 \right)$ with an initial velocity of $200$ feet Per second. At what instant will it be back at ground level?
A) $t = 10 \mathrm { Sec }$
B) $t = 12 \mathrm { Sec }$
C) $t = 11 \mathrm { Sec }$
D) $t = 9 \mathrm { Sec }$
E) $t = 20 \mathrm { Sec }$

Question

Use the position equation $s = - 20 t ^ { 2 } + v _ { 0 } t + s _ { 0 }$ , where $S$ represents the height of an object (in feet), $v _ { 0 }$ represents the initial velocity of the object (in feet per second), $s _ { 0 }$ represents the Initial height of the object (in feet), and $t$ represents the time (in seconds).
A projectile is fired straight upward from ground level $\left( s _ { 0 } = 0 \right)$ with an initial velocity of $200$ feet Per second. At what instant will it be back at ground level? 
A) $t = 10 \mathrm { Sec }$
B) $t = 12 \mathrm { Sec }$
C) $t = 11 \mathrm { Sec }$
D) $t = 9 \mathrm { Sec }$
E) $t = 20 \mathrm { Sec }$

Quizplus · Accepted Answer

To find when the projectile returns to ground level, set $s = 0$ and solve for $t$: $0 = -20t^2 + 200t$. Factoring out $t$, we get $t(-20t + 200) = 0$, giving $t = 0$ (launch time) and $t = 10$ seconds (return time).

Use the Position Equation s=−20t2+v0t+s0s = - 20 t ^ { 2 } + v _ { 0 } t + s _ { 0 }s=−20t2+v0​t+s0​

Use the Position Equation $s = - 20 t ^ { 2 } + v _ { 0 } t + s _ { 0 }$