An object is thrown into the air. Its height in feet above the ground t seconds later is given by h(t)=-16t^2+30t+25 a. Find the time when the object reaches its maximum hight b. How hih does it get
h = -16t2 + 30t + 25 height
v = -32t + 30 velocity
a. Maximum height occurs when v = 0; i.e. when 0 = -32t + 30; t = 30/32; t = 15/16 secs.
b. Height at this time: h = -16*(15/16)2 + 30*15/16 + 25 feet; h = 625/16 feet; h ≈ 39 feet
