Question

A baseball is thrown with a vertical velocity of 50 ft/s from an initial height of 6 ft. The height h in feet of the baseball can be modeled by h(t) = -16t2 + 50t + 6, where t is the time in seconds since the ball was thrown. It takes the ball approximately
seconds to reach its maximum height. (Round to the nearest tenth of a second and enter only the number.)

To the nearest foot, what is the maximum height that the ball reaches? (Enter only the number.)

A player hits a foul ball with an initial vertical velocity of 70 ft/s and an initial height of 5 feet. The maximum height reached by the ball is
feet. (Round to the nearest foot and enter only the number.)

I need the answers.

Answer 1

Vertex is highest point


ok so the t value of the vertex of a function in the form
f(x)=at^2+bt+c is
-b/2a

find t value
h(x)=-16t^2+50t+6
a=-16
b=50
-50/(2*-16)=1.5625
round
1.6 seconds
max height
input 1.6 for t and evaluate (for more precice, input 1.5625 for t)

max height is 45 feet



seems to be
b=initial velocity
c=intial height

a=-16
eqation is
h(t)=-16t^2+70t+5
vertex is -70/(2*-16)=2.1875
input for t to get max height
82 feet



answers are

1.6 seconds
45 feet
82 feet