An object is launched upward at 64 ft/sec from a platform that is 80 feet high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t² + 64t + 80? Round to the nearest foot.

Question

An object is launched upward at 64 ft/sec from a platform that is 80 feet high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t² + 64t + 80?  Round to the nearest foot.

anonymous · Answer

So the maximum height of an object will be atteined at the time when the velocity of the object goes to zero. The velocity is the derivative of positions so,$$v(t)=(-16t^2+64t+80) \prime = -32t+64=0$$ Solving for t, we see this happens at t=2seconds. Therefore, the height of the object at t=2 is:$$h(2)=-16*2^2+64*2+80=144feet$$

anonymous · Answer

so the answer is 144 feet