The height h, in feet, of a ball that is released 4 feet above the ground with an initial velocity of 80 feet per second is a function of the time t, in seconds, the ball is in the air and is given byh(t)=-16t^2+80t+4, 0 < t < 5.04a. Find the height of the ball above the ground 2 seconds after it is released.b. Find the height of the ball above the ground 4 seconds after it is released.Please show work!
Accepted Solution
A:
Answer with Step-by-step explanation:The height of the ball from the ground as a function of time is given by[tex]h(t)=-16t^2+80t+4[/tex]The height of the ball at any instant of time can be found by putting the value of time 't' in the above relation asPart a)Height of ball after 2 seconds it is released is [tex]h(2)=-16\times 2^2+80\times 2+4=100feet[/tex]Part b)Height of ball after 4 seconds it is released is [tex]h(4)=-16\times 4^2+80\times 4+4=68feet[/tex]