An object is launched upward at 45ft/sec from a platform that is 40 ft 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)=-16t^2+45t+40?
A: 65 ft
B: 71 ft
C: 80 ft
D: 93 ft

Question

An object is launched upward at 45ft/sec from a platform that is 40 ft 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)=-16t^2+45t+40?
A: 65 ft
B: 71 ft
C: 80 ft
D: 93 ft

abb0t · Answer

find the derivative, find the zero's and plug into h(t).

anonymous · Answer

I don't how to do that...

abb0t · Answer

Find the max, do you know how to do that?

agent0smith · Answer

Can you differentiate $$\Large h(t)=-16t^2+45t+40$$ ?

abb0t · Answer

x = $-\frac{b}{2a}$ where you have $ax^2+x+c$

abb0t · Answer

once you have your max, you can plug that into h(t) and also solve. Essentially, you get the same thing for your maximum.

abb0t · Answer

with or without differentiating.

anonymous · Answer

Thank you!