Question

Please help me out. :)

Find the initial velocity needed in order to shoot a projectile vertically upward a distance of 10,000 ft.

Answer 1

well, we can integrate gravity to get the folrmula if need be

Answer 2

am, can we use the differential calculus instead?

Answer 3

acceleration is constant:
a(t) = g

integrate to get velocity:
v(t) = gt + C, such that at t=0, C=Vi

v(t) = gt + Vi

integrate again to get position
h(t) = 1/2 gt^2 + Vi t + C, such that t=0, C=Hi (initial height)

h(t) = 1/2 gt^2 + Vi t + Hi

if Hi = 0, then we reduce it to:
h(t) = 1/2 gt^2 + Vi t

Answer 4

not sure if i could model that with diffy Qs, havent had a lot of practice

Answer 5

h(t) = 10000, and since its in feet, let g=-32

Answer 6

Ouch! :( what will I do?

Answer 7

10000 = -16t^2 + Vi t

0 = -16t^2 + Vi t - 10000, since the time it takes to get up to the top is the midpoint of the zeros ... its the axis of symmetry

Answer 8

\[t=\frac{-V_i}{-2(16)}\]
which can prolly be subbed back into the equation
\[0=-16\left(\frac{-V_i}{-2(16)}\right)^2+V_i\left(\frac{-V_i}{-2(16)}\right)-10000\]

Answer 9

solve for Vi ...

Answer 10

Wait.

Answer 11

how to solve that? :( sorry.

Answer 12

your asking a differential calculus questions and cant do basic algebra?

Answer 13

im pretty sure i dont need to show you how to add and multiply and cancel like terms and such ....

Answer 14

sorry.. wait. sorry again.

Answer 15

Vi^2 = -640000

Answer 16

the negative is off, but yeah

Answer 17

Vi = 800 rather. sorry.

Answer 18

\[0=-16\left(\frac{-V_i}{-2(16)}\right)^2+V_i\left(\frac{-V_i}{-2(16)}\right)-10000\]
\[0=-\frac{V_i^2}{2^2(16)}+\frac{V_i^2}{2(16)}-10000\]
\[0=-\frac{V_i^2}{4(16)}+\frac{2V_i^2}{4(16)}-10000\]
\[0=\frac{V_i^2}{4(16)}-10000\]
\[10000(4)(16)=V_i^2\]
\[100(2)(4)=V_i\]

Answer 19

so, if it need to go 800 ft/sec just to make it to 10000ft, we need to go at least 800 or faster

Answer 20

ahhh.. that's the final answer?