Ifaballisthrownverticallyupwardfromtheroofofa48foottallbuildingwithavelocityof112ft/sec,itsheightinfeetaftersecondsisS(t)=48+112t-16t^2.Whatisthemaximumheighttheballreaches?Whatisthevelocityoftheballwhenith

Question

If a ball is thrown vertically upward from the roof of a 48 foot tall building with a velocity of 112 ft/sec,its height in feet after seconds is S(t)=48+112t-16t^2.What is the maximum height the ball reaches?What is the velocity of the ball when it hits the ground (height 0

Answer 1

S(t)=48+112t-16t^2
=-16(t^2-7t+49/4)+244
=-16(t-7/2)^2+244
所以当t=7/2时有最大值244
速度函数为距离函数的微分即v=48/t+112-16t
当height=0时,S(t)=48
所以112t-16t^2=0.因为t!=0
所以t=7,即7秒后球才落地
所以此时的v=6.857