Question

A ball is thrown upward with a speed of 40 feet per second from the edge of a cliff 500 feet above the ground. What is the speed of the ball when it hits the ground? Use acceleration due to gravity as –32 feet per second squared and approximate your answer to 3 decimal places.

Answer 1

@zepdrix @IrishBoy123 @SithsAndGiggles

Answer 2

a(t)=-32
v(t)=-32t+C
v(0)=-32(0)+C
40=C
v(t)=-32t+40
What now?

Answer 3

Anti-Differentiate again to get your position function I suppose :)

Answer 4

Did you miss me? :D
Integrate it again. To get the position function, y(t).

Answer 5

Of course I did :p
Are we not looking for speed? Why would I need position, I did that at first and got an insanely high number.

Answer 6

And note that they gave you another piece of initial data for the new unknown constant that shows up: s(0)=500

Answer 7

You need to know at what time the object hits the ground,
then you can use that time to calculate velocity

Answer 8

I now have -16t^2+40t+00

Answer 9

@zepdrix so fast T.T

Answer 10

quickest, use \(v^2 = u^2 + 2ax\)

Answer 11

So you're looking for the time t when your object is at a height of 0.
That is when it hits the ground.

0=-16t^2+40t+500

Answer 12

I meant 500 not 00 btw

Answer 13

These guys and their fancy physics shortcuts :) lol

Answer 14

Eh I prefer calc over physics

Answer 15

Leme finish it up.

Answer 16

\[v^2-u^2=2gh,0^2-(40)^2=2*-32*h,h=\frac{ -40*40 }{ -2*32 }=25\]
so it goes 25 feet higher.
Now total distance =500+25=525 ft
initial speed u=o ft/s
using above formula again you can find v ,the speed when it touches the ground.

Answer 17

It is position.
500ft cliff above the ground=+500
Thrown upward with speed of 40 feet per second-it will eventually fall down to the ground which is negative=40t
acceleration is negative because velocity is negative
-32 feet per second squared
t is seconds
so second squared=-32t^2

Answer 18

Position s=0 means it is a at ground.
You find t to know the time it took to reach the ground.
You then used the velocity function by taking derivative of position.
Plug that time it reached the ground to the velocity function.

Answer 19

∫v(t)=s(t)=−16t2+40t+500

Answer 20

when it comes down g is positive
or g=32 ft/s^2

Answer 21

I got t=6.978

Answer 22

Since other solution is negative

Answer 23

well: \(\ddot x = a, \implies \dot x \dfrac{d \dot x}{dx} = a \implies \left|\dfrac{\dot x^2}{2} \right |_u^v = ax \) to meet all those calculus cravings!!

Answer 24

It is canceled correct?

Answer 25

Yall spewing out a bit too much enough

Answer 26

t=7? good

Answer 27

Is that the final answer or would I simply take t at 0 and plug it into v(t)?

Answer 28

it should be more than 40

Answer 29

Which was 7

Answer 30

correct, plug it in, v(7)=?

Answer 31

I am getting -184

Answer 32

It would be 184 ft/sec correct?

Answer 33

184 for `speed`
-184 is the velocity though

the negative tells us direction, that it's going down

Answer 34

I think the original question asked for velocity, ya?

Answer 35

Oh it said speed :) my bad

Answer 36

Nope speed

Answer 37

:)

Answer 38

Oh, you've got your units correct also, cool!

Answer 39

hehe

Answer 40

Thanks all

Answer 41

\[v^2-0^2=2*32*525,v=\sqrt{64*525}=8*5\sqrt{21}=40\sqrt{21}\]

Answer 42

@surjithayer +1 !!