A ball is thrown into the air with an upward velocity of 32 ft/s. Its height h in feet after t seconds is given by the function h = –16t2 + 32t + 6. a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. b. What is the ball’s maximum height?

5 months ago@Study2Learn

5 months agoI got 2 seconds, 6 feet, is that correct?

5 months agoYou may be right though, I don't want to get you wrong.

5 months agoI believe it would be 1 second, 22 ft though.

5 months ago@mhchen pls check this whenever you're online

5 months agocomplete the square: \(h(t) = –16t^2 + 32t + 6 \qquad \equiv \color{red} {22 - 16 (t-1)^2 } \) How big can that bit in red get?! And when does it get there?

5 months ago@TheSmartOne

4 months ago@TheSmartOne are you still there?

4 months agoYou just need to find the maximum. t = -b/2a That will give you the maximum x-coordinate which is the the time at which the ball is at maximum height Plugging in that t-value, and solving for h will give you the maximum height

4 months agoI did that before, and I got 2 seconds, 6 ft. is that correct?

4 months agoNo The equation is in the form of ax^2 + bx + c so do -b/2a You take the coefficients

4 months ago