Given that the height of the object is h(t)=-16t^2+96t a] <span>At which time(s) will it take the object be at a height of 128 feet? To solve this we proceed as follows: we need to solve for t at h(t)=128 thus -16t^2+96t=128 rewriting the above we get: -16t^2+96t-128=0 simplifying gives us: t^2-6t+8=0 factoring the quadratic we get: t^2-4t-2t+8=0 t(t-4)-2(t-4)=0 (t-2)(t-4)=0 thus t=2 and t=4 hence the object will be at height 128 at t=2 sec and t=4 sec
b]</span><span>How long will it take the object to return to the point of departure? </span>Here shall proceed as follows: At the point of the departure, h(t)=0 thus plugging this in the equation we get: -16t^2+96t=0 simplifying this we get: -16t^2=-96t dividing through by -16 we get t^2=6t hence t=6 sec The time taken for the object to return to the departure is 6 sec
c] <span>How long will it take the object to reach its maximum height? </span>Time taken to reach the maximum height will be calculated as follows: h(t)=-16t^2+96t at maximum height h'(t)=0 but h'(t)=-32t+96 equating this to zero we get: -32t=-96 thus t=3 sec thus time taken to reach the height will be t=3 sec
d]<span>What is the Maximum Height? To get the maximum height we plug t=3 sec and simplify. h(t)=-16t^2+96t h(3)=-16(3)^2+96(3) h(3)=-48+288 h(3)=240 ft</span>
