s= -16t^2+100t

Question

I don't known where to start

Answer 1

This is the equation for the height (s) of an object that is propelled upwards with a velocity = 100 feet/sec. The -16t² term is due to the downward force of gravity. You didn't tell us what you are asked to determine, so I'm guessing you want to know when the object will reach maximum height and when it will hit the ground.

 

First, note that this is a quadratic equation, so the graph is a parabola.  Because the coefficient of the t² term is negative (-16), it is an inverted (upside down) parabola with the vertex at the top.  The vertex is always located at the coordinate t = -b/2a, where a is the coefficient of the t² term (-16) and b is the coefficient of the t term (100).  So the object reaches its maximum height at time:

 

     t = -100/2*(-16) = 100/32 = 3.125 sec.

 

To figure out what that height is, plug t = 3.125 into the s equation:

 

     s = -16*3.125² + 100*3.125 = ______ feet

 

When the object hits the ground, its height will be zero; that is, s = 0.

 

     s = -16t² + 100t

     0 = -16t² + 100t

     16t² = 100t

     16t = 100

     t = 100/16 = 6.25 sec.