a projectile is fired straight up from ground level with an initial velocity of 112 ft/s. its height, h…
a projectile is fired straight up from ground level with an initial velocity of 112 ft/s. its height, h, above the ground after t seconds is given by h = -16t² + 112t. what is the interval of time during which the projectiles height exceeds 192 feet?\no 3 < t < 4\no t < 4\no t > 4\no 3 > t > 4
Answer
Explanation:
Step1: Set up the inequality
We want to find when $h>192$, so we set up the inequality $- 16t^{2}+112t>192$. Rearrange it to the standard - quadratic form $16t^{2}-112t + 192<0$. Divide through by 16 to simplify: $t^{2}-7t + 12<0$.
Step2: Factor the quadratic
Factor the quadratic expression $t^{2}-7t + 12$. We get $(t - 3)(t - 4)<0$.
Step3: Find the roots
Set $(t - 3)(t - 4)=0$. The roots are $t = 3$ and $t = 4$.
Step4: Test intervals
We consider the intervals $t<3$, $3<t<4$, and $t>4$. For $t = 2$ (in the interval $t<3$), $(2 - 3)(2 - 4)=(-1)\times(-2)=2>0$. For $t = 3.5$ (in the interval $3<t<4$), $(3.5 - 3)(3.5 - 4)=(0.5)\times(-0.5)=-0.25<0$. For $t = 5$ (in the interval $t>4$), $(5 - 3)(5 - 4)=2\times1 = 2>0$.
Answer:
A. $3<t<4$