A projectile is fired straight upward from the ground level with an initial velocity of $112$ feet per second, so that its height h at any time t is given by $h = - 16 t ^ { 2 } + 112 t$ , where h is measured in feet and t is measured in seconds. During what interval(s)of time with the height of the projectile exceed $160$ feet?
A) $( 2,5 )$
B) $( 2,7 )$
C) $( 5,7 )$
D) $( 0,2 ) , ( 5,7 )$
E) $( 1,7 )$

Question

Quizplus · Accepted Answer

To find when the height exceeds 160 feet, set the equation $h = -16t^2 + 112t > 160$. Rearranging gives $16t^2 - 112t + 160 < 0$. Factoring or using the quadratic formula, we find the roots of the equation $16t^2 - 112t + 160 = 0$ are $t = 2$ and $t = 5$. The inequality is satisfied for $t$ in the interval $(2, 5)$, meaning the projectile's height exceeds 160 feet between 2 and 5 seconds after launch.

A Projectile Is Fired Straight Upward from the Ground Level 112112112

A Projectile Is Fired Straight Upward from the Ground Level $112$