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

when u shoot an object vertically up, only force acting on it is gravity

Answer 2

\(a = -9.8\)

Answer 3

u wanto use kinematics equations or calculus ?

Answer 4

Calculus please. :)

Answer 5

is a referring to ACCELERATION? :)

Answer 6

yup !

Answer 7

ahh. :)

Answer 8

\[a = -10 \]


\[v(t)-v(0) = \int \limits_0^t a(t) dt = \int \limits_0^t-10~ dt \]

Answer 9

is it integral calculus?

Answer 10

\[v(t)-v(0) = \int \limits_0^t a(t) dt = -10t \]

\[v(t) = -10t + v(0) \]

Answer 11

yes, you're not familiar wid integrals yet ?

Answer 12

familiar.. but only the basic.

Answer 13

is it possible to use the differential calculus?

Answer 14

yes

Answer 15

yehey! can we use that one please? :)

Answer 16

seen this equation before :

\(s = ut + \dfrac{1}{2}at^2\)

?

Answer 17

not yet. :) just now. :)

Answer 18

ohk, then i would simply use kinematics equation :

\(v^2 - u^2 = 2as\)

at max height, velocity = 0 :

\(0 - u^2 = 2as\)

Answer 19

plugin a = -9.8, s = 10000
and solve \(u\)

Answer 20

http://www.wolframalpha.com/input/?i=solve+0+-+u%5E2+%3D+-2*9.8*10000*0.3048

Answer 21

@ganeshie8 i think:
\[h =h_0+V_0t +at^2\] \(h_0=0\) so that \(h= V_0t+at^2\) at the required height h = 10000, we have 10000=\(V_0t +at^2\)
let it there
now, \(V= V_0+1/2 at\) at the maximum height, V=0 so that \(V_0= -1/2 at^2\)
replace it back to the equation above to find t, first, when we have t, replace back to find \(V_0\)

Answer 22

@Loser66 \(V = V_0 + at \) right ?

Answer 23

1/2 at, take derivative of h

Answer 24

what h are u talkign about ?

Answer 25

we know that derivative of distance is velocity, right?
h = h0+ V0t +at^2
dh/dt = V = V0 +1/2at

Answer 26

okay, derivative of h gives u
dh/dt = V0 + 2at

Answer 27

oh yaa. hahaha. I am sorry.

Answer 28

but thats incorrect still

Answer 29

the starting equation for h is wrong unless u define a = 2g

Answer 30

\[a = -g\]

\[v(t) = -gt + v(0) \]

\[h(t) = -g\frac{t^2}{2} + v(0)t + h(0) \]

Answer 31

integrating the constant acceleration backwards two times gives u above equation for height , right ?

Answer 32

oh, let me redo.
\[h = V_0t + at^2\]
\[V_0 = -2at\]
\[10000= -2at^2+at^2 = -at^2 \] a =-10
so t=100
\[V_0 = -2(-10)100= 2000

Answer 33

doesnt look correct

Answer 34

hey! im sorry. :( I've lost the connection.. :(
Can you teach me again?

Answer 35

starting equation for h is incorrect, since you're using a = -10

Answer 36

and one more mistake is the given height is in feet, not meters... and the acceleration you're using is in meters/sec^2... so ..

Answer 37

Ok, I am sorry if it is not correct.
@ganeshie8 don't bother by -10 or 10 on a . We construct the formula and work on the letter to get the last equation. After all, plug the final numbers in
why do we have to plug them in too soon and then confuse anything?

Answer 38

Agree !

Answer 39

if it is just unit, so?? convert the unit, done, right?

Answer 40

yup !

Answer 41

I am sorry, I have to go. :) you help the Asker, Ok ? my friend?