Ifaballisthrownverticallyupwardfromtheroofofa48foottallbuildingwithavelocityof48ft/sec,itsheightinfeetaftertsecondsiss(t)=48+48t-16t^2.Whatisthemaximumheighttheballreaches?Whatisthevelocityoftheballw

Question

If a ball is thrown vertically upward from the roof of a 48 foot tall building with a velocity of 48 ft/sec,its height in feet after t seconds is s(t) = 48 + 48 t - 16 t^2.What is the maximum height the ball reaches?
What is the velocity of the ball when it hits the ground (height 0)?
第一个问题答案是84,但第二个的没有人对呢...

Answer 1

s(t) = -(t-3/2)^2+108
所以当t=3/2的时候s(t)达到最高值108
速度也就是距离对时间的变化率,可以通过求导来做
s(t)'=-32t+48
当s(t)=0的时候,通过第一个等式可以求得t=6√3+3/2
代入s(t)' 得最终速度是 -192√3 （负号代表方向）