Question

The height h (in feet) of an object shot into the air from a tall building is given by the function h(t)=800+80t-16t^2, where t is the time elapsed in seconds. What should be the velocity when it reaches its highest point? when does this occur? And what is the maximum height reached by the object?

Answer 1

i know the velocity is v(t)=80-32t

Answer 2

I just don't know where to go from there.

Answer 3

velocity should be zero as the object is stationary for an instant before it begins to fall

Answer 4

the highest point is at the max value of the height functn

Answer 5

ur derivative of the height function appears correct

Answer 6

ok.

Answer 7

so different s(t) wrt t, and find the value of time from there by equating it to zero

Answer 8

so I would just set the velocity equal to zero, solve for x then plug it into the position function?

Answer 9

yes..that gives you the maximum height

Answer 10

do you think its appropriate to answer the second question as "right before the ball begins to fall"?

Answer 11

yes

Answer 12

ok wonderful! thank you!

Answer 13

Set \(80-32t=0 \implies t=\frac{5}{2}\). So, at \(t=\frac{5}{2}\), the object reaches its maximum hieght. Its velocity at that point must be \(0\) and its hieght is \(800+80\frac{5}{2}-16(\frac{5}{2})^2\).

Answer 14

thanks guys.

Answer 15

i have to put my two cents in sorry. you do not need any calculus for this. the vertex of a quadratic polynomial is given by \[-\frac{b}{2a}\] and in this case you get
\[\frac{80}{32}=\frac{5}{2}\] just thought i would mention it now i will be quiet

Answer 16

thanks guys.