Question

a ball is thrown vertically upwards from a height of 6 ft with an initial velocity of 40 ft per second. how high will the ball go?

Answer 1

well if we stay in ft and since we do not know a time (yet we can find this out without a time) we use the function

\[v'^{2}=v^{2}-2gy\]

We want to know y. We know g is 32ft/s^2, initial velocity (v) is 40 ft/s and at the top of the projection (max height) our final velocity (v') is zero correct?

Answer 2

correct alll it wants is the max hieght

Answer 3

is this for a calculus class on integration?

Answer 4

yea its a calculus question

Answer 5

if so, then we have \(\int g \ dt\) where \(g\) is a constant for gravity, and we also know

\(v(0)=40\)

and \(d(0)=6\)

Answer 6

correct so insert the values

\[0=200ft^{2}/s^{2}-2(32ft/s^{2})y\]
\[-200ft^{2}/s^{2}=(-64ft/s^{2})y\]
\[y=3.125ft\]

Answer 7

and yea we need to include the initial height of 6ft so 9.125ft

Answer 8

well i have possible answers and it doesn't have 3.125 ft

Answer 9

here ill list the answers a. 81 ft, 24.7 ft c.31 ft d. 69 ft e. 84 ft

Answer 10

\(v(t) = \int g \ dt= gt+c\\
v(0) = 40 = c\\v(t)=gt+40\\p(t)=\int v(t) dt=\frac{1}{2}gt^2+40t+c_0\\p(0)=6=c

\\p(t)=\frac{1}{2}gt+40t+6\)

The vertex \(t\) coordinate of the vertex is given by \(t=\frac{-40}{\frac{g}{2}}\)

Plug in g for gravity (it should be negative) and then plug the result into \(p(t)\)

Answer 11

What are you using for gravity?

Answer 12

well the questions doesn't really give out the gravity

Answer 13

That should say \(t=\frac{-40}{2(g\frac{1}{2})}=\frac{-40}{g}\)

we will assume g = is -32 (negative because its working against us)

\(p(\frac{-40}{-32})=p(\frac{5}{4})=31ft/s\)

Answer 14

oh k, thanks zzr0ck3r for the help!

Answer 15

31 ft

not ft/sec...