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 ca
n 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.)
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
