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 :
[tex]h=-16t^2+112t[/tex]
We need to find the interval of time during which the projectile’s height exceeds 192 feet. It means put h = 192 feet
So,
[tex]-16t^2+112t=192\\\\-16t^2+112t-192=0[/tex]
The above is a quadratic equation, it can be solved using middle term splitting as follow :
[tex]16 (t^2 - 7t + 12) = 0\\\\16 (t - 4)(t - 3) = 0\\\\t=4\ s\ \text{and}\ t=3\ s[/tex]
Hence, at t = 3 s and t = 4 s, the projectile’s height exceeds 192 feet above the ground.
