A ball is thrown vertically upward from the top of a building 160 feet tall with an initial velocity of 48 feet per second. The distance d (in feet) of the ball from the ground after t seconds is d(t) = 160 + 48t – 16t2. a. After how many seconds does the ball strike the ground? Write your answer in a complete sentence with proper grammar and correct spelling. b. When will the ball reach its maximum height?

Answergiven,height of the building = 160 ftinitial velocity = 48 ft/sthe relation of distance is givend = -16 t²  + 48 t + 160time at which the ball hit the ground will be when d = 0-16 t²  + 48 t + 160 = 0 solving quadratic equation[tex]t = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex][tex]t = \dfrac{-48\pm \sqrt{48^2+4\times 16 \times 160}}{2\times (-16)}[/tex][tex]t = \dfrac{-48\pm 112}{-32}[/tex][tex]t = \dfrac{-48+ 112}{-32},\dfrac{-48- 112}{-32}[/tex]t = 5 s(neglecting the negative solution)b) ball will reach max height when velocity is equal to zero   [tex]v = \dfrac{dd}{dt}[/tex]   [tex]v = \dfrac{d}{dt}(-16 t^2+48 t +160)[/tex]   [tex]v = -32 t + 48[/tex]     v = 0      t = 1.5 s at 1.5 sec the ball will reach at maximum height