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.

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.

anonymous · Answer

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.)

anonymous · Answer

The first two I believe are: 162.5 and 311.25

anonymous · Answer

$$h(t) = -16t^2 + 50t + 6$$ the max height is achieved when $t=-\frac{b}{2a}=\frac{50}{32}$

anonymous · Answer

so no, not $162.5$

anonymous · Answer

To the nearest foot, what is the maximum height that the ball reaches
find 
$$h(\frac{25}{16})$$ i would use a calculator