Use the position equation $s = - 16 t ^ { 2 } + v _ { 0 } t + s _ { 0 }$ ,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 176 feet per second.At what instant will it be back at ground level?

A) $t = 11$ Sec
B) $t = 16$ Sec
C) $t = 13$ Sec
D) $t = 12$ Sec
E) $t = 10$ Sec

Question

Use the position equation $s = - 16 t ^ { 2 } + v _ { 0 } t + s _ { 0 }$ ,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 176 feet per second.At what instant will it be back at ground level?
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A) $t = 11$ Sec
B) $t = 16$ Sec
C) $t = 13$ Sec
D) $t = 12$ Sec
E) $t = 10$ Sec

Quizplus · Accepted Answer

To find when the projectile will be back at ground level, set $s = 0$ and solve for $t$ using the given equation $s = -16t^2 + v_0t + s_0$. Given $v_0 = 176$ feet per second and $s_0 = 0$, the equation becomes $0 = -16t^2 + 176t$. Factoring out $t$, we get $t(-16t + 176) = 0$, which gives $t = 0$ (the initial time) and $t = 11$ when solving $-16t + 176 = 0$. Thus, the projectile will be back at ground level at $t = 11$ seconds.

Use the Position Equation s=−16t2+v0t+s0s = - 16 t ^ { 2 } + v _ { 0 } t + s _ { 0 }s=−16t2+v0​t+s0​

Use the Position Equation $s = - 16 t ^ { 2 } + v _ { 0 } t + s _ { 0 }$