Graph the quadratic equation from t=0 to the value of t for which the ball hits the ground.

Question

A person standing close to edge on the top of a 200-foot building throws a baseball vertically upward. The quadratic function s(t)=-16t^2+64+200 models the balls height above the ground, s(t), in feet, t seconds after it was thrown

Answer 1

Allow me to correct your equation.

 

s(t) = -16t² + 64t + 200

 

 

When graphing a parabola, we need several points:

 

x-intercepts (in this case, ground level)

y-intercept (in this case, initial height of the ball)

Vertex (in this case, maximum height of the ball)

 

 

To find the y-intercept, set t=0 and solve for s(t).

 

s(0) = 200

 

y-intercept = (0, 200)

 

 

To find the x-intercepts, set s(t)=0 and solve for t.

 

16t² + 64t + 200 = 0

 

 

Use the quadratic formula to solve for t:

 

t = (-b ± √(b² - 4ac)) / 2a

 

where:

a = -16

b = 64

c = 200

 

Plug in these values into the formula.  You will have 2 t values.  Due to the restriction of t in your question, you only accept the positive t value.

 

The x-intercept is then (t, 0).  x is the value you chose.  It is also the point that represent ground level.

 

 

To find the vertex, use the formula

 

t = -b/2a  to find the time in which the ball reaches its maximum height.

 

 

t = -64 / (2 * -16)

t = 2

 

 

Evaluate  s(2) to find the vertex.

 

s(2) = -16(2)² + 64(2) + 200

       = -64 + 128 + 200

       = 264

 

 

Vertex is (2, 264).

 

 

Finally, plot all these points on a coordinate system then connect them so that you get a parabola.  Your parabola should not go below the x-axis, since the x-axis represents ground level.