Question

Physics stuff!!

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

@ganeshie8 @nincompoop @pooja195 @Preetha any ideas?

Answer 2

I believe there are two ways to do this

1) Calculate the trajectory

2) Calculate the energy change

Answer 3

presuming the ball lands on the ground and not on the cliff... even though it was thrown vertically upwards

Answer 4

this is a calc I problem by the way

Answer 5

maybe the angle shouldn't matter here because the ball gains its speed by the time it reaches cliff. You may start with the given acceleration due to gravity :

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

Answer 6

okay s v(0) = 40 but how do we find v(t) ?

Answer 7

Continue further,

\[v(t) = \frac{d x(t)}{dt}\]

Answer 8

wait slow down , how did you find s(t) ?

Answer 9

\[\begin{align}v(t)-v(0)&=\int\limits_0^t a(t)\,dt \\~\\
&=\int\limits_0^t -32\,dt
\end{align}\]

you're given that \(a(t)=-32\), see if you can integrate above now..

Answer 10

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 11

v(t) - 40 = (-32t^2)/2 ?

Answer 12

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 13

nope, whats antiderivative of constant ?

Answer 14

well you add a variable in this case t

Answer 15

oh woops i see what i did wrong

Answer 16

v(t) = -32t + 40:

also @Shalante if we take your s(t) and we derive it until we get a(t) we get -64 and i think it should be -32

Answer 17

no it is -64

Answer 18

\[\int\limits v(t) = s(t) = -16t^2 + 40t + 500\] does this look right to you ?

Answer 19

"se acceleration due to gravity as –32 feet per second squared"

Answer 20

use *

Answer 21

Actually, there lots of ways to do this.
There is like 3

Answer 22

yes that is right.

Answer 23

i know but i have to show my work for this one so cant use physics equations sadly :(

Answer 24

oh wait it is wrong. Supposed to be -32t^2-40t+500

Answer 25

Make the position=0 by doing quadratic formula.

Answer 26

use your calculator
divide everything by -4 first

Answer 27

Yeah as far as physics is concerned I think this would be pretty much the best way:

\[K_i + V_i = K_f + V_f\]
\[\frac{1}{2} m 40^2 + m 32*500 = \frac{1}{2} m v^2\]
\[\sqrt{40^2 + 2*32*500} = v\]

Answer 28

careful, indefinite integral is unique only upto a constant

Answer 29

interesting, i'm getting positive 40t because of the v(t) - v(0) part (when we added the v(0) to isolate the v(t))

Answer 30

\[s(t)-s(0)=\int\limits_0^t v(t) -16t^2 + 40t \]

so you're right, you do get \(s(t) = -16t^2+40t+500\)

Answer 31

okay so now just set s(t) = 0 to find when it touches the ground right?

Answer 32

wait i thought s (t) = -16t^2 .... not -32t^2

Answer 33

physicists are overjoyed by the conservation of "anything", but i think math professors still want to see the integrals mess ;)

Answer 34

Answer is negative because the ball's speed while hitting the ground is going downwards.

Answer 35

t=0 corresponds to s=500 right ?

Answer 36

recall that \(s(t)\) is the displacement from \(ground\) after \(t\) seconds

Answer 37

\[-32t^2+40t+500=0 \]
got \[t=3.376s,-4.626s\]
Time should only be positive so use t=3.376s
\[v(t)=-64t-40\]
\[v(3.376)=-256.06 m/s\]

are you sure this is right @ganeshie8

Answer 38

shouldn't it be \[-16t^2 +40t + 500 = 0 \]

Answer 39

sry im not following the discussion between you and Jdosio... but you're not allowed to use direct formulas here right ?

Answer 40

no only calculus

Answer 41

okay my only positive root for v(t) is 6.978

Answer 42

http://www.wolframalpha.com/input/?i=-16x%5E2+%2B+40x+%2B+500

Answer 43

Ya use -16t^2

Answer 44

so to find the velocity at that time we just plug that into the formula for v(t) right ?

Answer 45

which gives me around -183.3030

Answer 46

does that look about right to you guys?

Answer 47

Nope it should be -40

Answer 48

im using v(t) = -32t + 40 is that right?

Answer 49

Answer is right, but the graph is wrong.
Guess you do not need the graph.

Answer 50

okay so then does everyone agree that the answer is -183.3030 !?

Answer 51

Read it wrong at the beginning lol

Answer 52

@Empty @ganeshie8 ?

Answer 53

It is correct.

Answer 54

okay ill stick around just incase anyone changes their mind, thanks for the help guys :D

Answer 55

I got -183.3030 by doing it with the energy considerations alone, so good job! Same answer different route is a good indicator imo.

Answer 56

Yes the ball touches the ground after at `t=6.978`

\(v(6.978) =40-32(6.978) \approx -183.303\)


you want to avoid that negative sign though
speed is \(183.303\) whatever unit

Answer 57

Only reason why I think they'd want the negative sign there is because they gave -32 f/s^2 as the acceleration due to gravity, which imo is kind of awkward so I don't know what they want.

Answer 58

acceleration is a vector, so sign makes sense
speed isn't

Answer 59

ill just make it explicit that the velocity is downwards

Answer 60

they are not asking velocity

Answer 61

Yeah you're right @ganeshie8 good point.

Answer 62

just persisting because, when you use the terms correctly, your professor knows that you understood the concepts..

that is for @Jdosio :)

Answer 63

ill take note of that :)

Answer 64

okay something weird has happened

Answer 65

i found the speed of the ball at the time that the ball is at a height of 500 (the height of the cliff) and got -40 does that look oddly round to you guys ?

Answer 66

@ganeshie8 @Empty @Shalante ?

Answer 67

Well this is actually completely expected that the speed is the same for the same height, this is because it has no more potential or kinetic energy at this same height when you threw it. This is due to conservation of energy, which I realize is not exactly in the scope of what you're learning however it's clear at a glance to me because of this.

Answer 68

okay so i should stick with my 183.3030 right?