Question

A projectile is fired straight up from ground level with an initial velocity of 112ft/s. It’s height, h, above the ground after t seconds is given by h=-16t2+112t. What is the interval of time during which the projectile’s height exceeds 192 feet?

Answer 1

Answer:

t = 3 s and t = 4 s

Step-by-step explanation:

Given that,

A projectile is fired straight up from ground level with an initial velocity of 112ft/s. Its height as a function of time t is given by :

$h=-16t^2+112t$

We need to find the interval of time during which the projectile’s height exceeds 192 feet. It means put h = 192 feet

So,

$-16t^2+112t=192\\\\-16t^2+112t-192=0$

The above is a quadratic equation, it can be solved using middle term splitting as follow :

$16 (t^2 - 7t + 12) = 0\\\\16 (t - 4)(t - 3) = 0\\\\t=4\ s\ \text{and}\ t=3\ s$

Hence, at t = 3 s and t = 4 s, the projectile’s height exceeds 192 feet above the ground.