+83 The equation $y = -16t^2 - 18t + 405$ describes the height (in feet) of a ball thrown downward at 18 feet per second from a height of 405 feet from the ground, as a function of time $t$, in seconds. In how many seconds will the ball hit the ground? Express your answer as a decimal rounded to the nearest tenth.
Thanks!
Don't know where to start.
+129852 When the ball hits the ground the height = 0
Multiplying the original equation through by -1 and setting this to 0 we have
16t^2 + 18t - 405 = 0
Use the Q Formula
t = ( -18 ± sqrt [ 18^2 + 4*16 *405 ] ) / [ 2 * 16 ] = [ -18 ± 162 ] / 32
Take the positive root
t = [ -18 + 162 ] /32 = 144 / 32 = 4.5 sec
