Question

The trajectory of an object that is thrown from the top of the building with an initial velocity of 15 m/s is modeled as H equals -4.9 T to the second power +52+ X, where HIs the height in meters of the object at a given time T is the time in seconds of X is the height of the building in meters. If the object hits the ground after 21 seconds, what is the approximate height in meters of the building

Answer 1

Answer:

Step-by-step explanation:

This is our quadratic:

$s(t)=-4.9t^2+15t+h_{i}$

where s(t) is the position of the object after all the stuff on the right side happens to it.  If the object lands on the ground after all that stuff happens to it, its height is 0.  Subbing in a 0 for s(t) gives us:

$0=-4.9t^2+15t+h_{i}$

and if the object hits the ground at 21 seconds, we can sub in a 21 for each t to get this:

$0=-4.9(21)^2+15(21)+h_{i}$ and do the math.

$0=-1845.9+h_{i}$ and solving for the height:

h = 1845.9 meters