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 projectile's height exceeds 192 feet?

Respuesta :

Answer:

Step-by-step explanation:

We can do this the easy way and just set up an inequality and let the factoring do the work for us. The inequality will look like this:

[tex]-16t^2+112t>192[/tex] We will move the constant over and get

[tex]-16t^2+112t-192>0[/tex] and when you factor this you get that

3 < t < 4

Between 3 and 4 seconds is where the projectile reaches a height higher than 192 feet. With a little more work and some calculus you can find the max height to be 196 feet.

Answer:

its A

Step-by-step explanation:

got it right

ACCESS MORE
EDU ACCESS
Universidad de Mexico