Question

An object is projected upward from ground level with an initial velocity of 660 feet per second. (In this exercise, the goal is to analyze the motion of the object during its upward flight.)

(a) If air resistance is neglected, find the velocity of the object as a function of time. Use a graphing utility to graph this function.


(b) Use the result of part (a) to find the position function.

Determine the maximum height attained by the object.

Answer 1

Calculus?

Answer 2

\[
v = \frac{dy}{dt} = u - gt \\
y = \int (u - gt)dt
\]u = 660 ft / sec
g = 32 ft / sec^2

Answer 3

(c) If the air resistance is proportional to the square of the velocity, you obtain the equation

\[\int\limits \frac{ dv }{ 32+kv^2 } = -\int\limits dt\]

where −32 feet per second is the acceleration due to gravity and k is a constant. Find the velocity as a function of time by solving the equation

v(t)= _______________________

(d) Use a graphing utility to graph the velocity function v(t) in part (c) for k = 0.003. Use the graph to approximate the time t0 at which the object reaches its maximum height. (Round your answer to two decimal places.)

\[t _{0} = ______ s\]


(e) Use the integration capabilities of a graphing utility to approximate the integral
\[\int\limits_{{0}}^{t_0}v(t) dt\]

where v(t) and \[t_{0}\] are those found in part (d). This is the approximation of the maximum height of the object. (Round your answer to two decimal places.)

h = ______ ft

Answer 4

Calculus II, not I. ._. @aum

Answer 5

I know for the (a) that v(t) = -32t+660

Answer 6

And that for (b) that the position function is \[-16t^2+660t\]

Answer 7

a) \(\large v = u - gt \)

\(\large v = 660 - 32t\)

Answer 8

I'm stuck on how to find maximum height... ._.

Answer 9

Well maximum height of the object...

Answer 10

yes, b) \(\large y = -16t^2+660t \)
This is a parabola and we need to find the y-coordinate of the vertex.
The x-coordinate of the vertex is given by -b/(2a) (for ax^2 + bx + c)
-b/2a = -660 / (2 * -16) = ? seconds.
put that t in \(\large y = -16t^2+660t \) and find \(\large y_{max}\).

Answer 11

Doing that now.

Answer 12

6806.25= y max

Answer 13

Yes, maximum height = 6806.25 feet.

Max. height can also be found this way: When max height is reached, velocity will be zero because if the velocity is not zero the object will keep moving upwards. Since it has stopped and about to return, v = 0 at max height.
From a) v = 660 - 32t
0 = 660 - 32t
t = 20.625 sec.
ymax = -16(20.625)^2 + 660(20.625) = 6806.25 feet.

Answer 14

Oh, wow! Got it. :) That makes that part easier.

Answer 15

For c) make a tan substitution. Or use the standard integral for
\(\large \int \frac{dx}{a^2+x^2}\)

Answer 16

The substution and integration I'm having trouble with. Rusty skills, but I'm trying to work on them.

Answer 17

But for the left side of the equation, the tan substitution can be applied right? I'm trying to understand. :)

Answer 18

\[
\int\limits \frac{ dv }{ 32+kv^2 } = -\int\limits dt \\
\frac 1k \int\limits \frac{ dv }{ 32/k+v^2 } = -\int\limits dt \\
\frac 1k \int\limits \frac{ dv }{ \left(\sqrt{32/k}\right )^2+v^2 } = -\int\limits dt \\
v = \left(\sqrt{32/k}\right )\tan(t)
\]
Or you can look up standard integration table for \[ \int\limits \frac{ dx }{ a^2+x^2 }\]

Answer 19

\[\int\limits \frac{ dx }{ a^2+x^2 } = \frac 1a \tan^{-1}\frac xa + C\]

Answer 20

gtg.

Answer 21

Okay, thank you for the help though. :)