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Physics 11 Online
OpenStudy (anonymous):

A cannonball is shot at an angle of 45 degrees to the horizon with a velocity of 100m/s. a. how high above the ground will it get? b. how far away from the cannon will it hit the ground? c. at what velocity will it hit the ground?

OpenStudy (anonymous):

First step is to use vectors to determine the vertical (y) component of the cannon balls acceleration I can draw a diagram if you want

OpenStudy (anonymous):

yes please

OpenStudy (anonymous):

sweet gimme a second

OpenStudy (anonymous):

|dw:1382324546978:dw|

OpenStudy (anonymous):

How do you think you would find Y?

OpenStudy (anonymous):

you take the sin of 45 degrees and multiply by 100 right?

OpenStudy (anonymous):

Yup! now we have the vertical component of velocity, we can find how high it will get using kinematic equations Are you familiar with kinematic equations?

OpenStudy (anonymous):

not really haha

OpenStudy (anonymous):

Sweet as, we know the initial velocity, the acceleration of the cannonball (which is the acceleration of gravity 10 ms^-2), the final velocity ( which is zero at its highest point) and want to find the distance traveled. we can look at some kinematic equations, and find one that has those terms, then use it to find distance. http://www.physicsclassroom.com/class/vectors/u3l2e3.gif look at these and find one that has, initial velocity, final velocity, acceleration and displacement

OpenStudy (anonymous):

the 3rd equattion

OpenStudy (anonymous):

Nice! so first off lets just get rid of Vf^2, as we know its zero haha this give us a nice equation 0 = Vi^2 + 2ad rearranging for d ( which is the distance we want) we get -Vi^2 / 2a = d

OpenStudy (anonymous):

now just plug in Vi and a, and you will get d, which is the vertical distance the cannonball will travel

OpenStudy (anonymous):

vi is 100 and acceleration is -9.81 right?

OpenStudy (anonymous):

sure is

OpenStudy (anonymous):

so its 509.68 ?

OpenStudy (anonymous):

yup well done

OpenStudy (anonymous):

For the second question we need the time that that ball will be in the air, so find an equation from that page i linked you that gives us the time, relative to acceleration and velocity.

OpenStudy (anonymous):

wait so which part did we just answer?

OpenStudy (anonymous):

Part a! we found the vertical distance that the cannonball with travel

OpenStudy (anonymous):

oh ok gotcha haha!

OpenStudy (anonymous):

and the second equation right?

OpenStudy (anonymous):

So now we want to find the distance traveled in the x direction

OpenStudy (anonymous):

yup the second equation is correct

OpenStudy (anonymous):

so vf=vi+a*t? vf is 100 a is -9.81 and we have to solve for t still?

OpenStudy (anonymous):

we sure do. that gives us half the time that the ball spends in the air, so the time the ball takes to get to its maximum height

OpenStudy (anonymous):

we double time

OpenStudy (anonymous):

yup, if we double time, it gives the the entire time that the ball is in the air, which means we can calculate the distance traveled in the x direction during that time using the first equation

OpenStudy (anonymous):

i got t=-10.19

OpenStudy (anonymous):

yup, except time is positive in this case, so we can drop the negative sign to get 10.19 seconds

OpenStudy (anonymous):

so thats the time for the ball to get to its highest point, and it will take the same amount of time to get back down again, so double that time for the total flight time!

OpenStudy (anonymous):

20.38?

OpenStudy (anonymous):

yup, so what else do we need for the first equation that we dont already have? (its only one more thing)

OpenStudy (anonymous):

vi and thats 0 right?

OpenStudy (anonymous):

we do need Vi, but Vi is not zero

OpenStudy (anonymous):

Think of how we found Vi for the first question

OpenStudy (anonymous):

ooooooh so its cos(45)x100

OpenStudy (anonymous):

and thats 85.09

OpenStudy (anonymous):

yup! well done!

OpenStudy (anonymous):

thank you!

OpenStudy (anonymous):

no problem, now plug all your values into the first equation to find the distance traveled

OpenStudy (anonymous):

do i still square the time like how it says in the last part or do i multiply time by 2

OpenStudy (anonymous):

always just do what the equation tells you, so in this case square time. We have already doubled time to get 20.38 seconds remember. What value are you using for a? ( acceleration)

OpenStudy (anonymous):

-9.81?

OpenStudy (anonymous):

What direction does gravity work?

OpenStudy (anonymous):

x or y?

OpenStudy (anonymous):

y...

OpenStudy (anonymous):

i got d=357.75m

OpenStudy (anonymous):

So if gravity works in the y direction, do you think that it would have any effect on the distance traveled in the x direction?

OpenStudy (anonymous):

remember we are trying to find the distance traveled in the x direction

OpenStudy (anonymous):

well yes because if it wasnt it would just keep going? a force has to bring it down right?

OpenStudy (anonymous):

Gravity determines the time the ball is in the air, but has no effect on the horizontal velocity

OpenStudy (anonymous):

As there are no forces acting on the ball in the x direction, there is no acceleration of the ball in the x direction. For the second question acceleration is therefore zero.

OpenStudy (anonymous):

its asking how far from the cannon will it hit the ground, so gravity doesnt matter here?

OpenStudy (anonymous):

we are working in the x direction, gravity only effects y direction

OpenStudy (anonymous):

so its just d= (85.09)(10.19)+1/2(0)(10.19)^2?

OpenStudy (anonymous):

yup sure is

OpenStudy (anonymous):

except time is double remember

OpenStudy (anonymous):

so t = 20.38

OpenStudy (anonymous):

d=1734.1342?

OpenStudy (anonymous):

Nicely done

OpenStudy (anonymous):

seems like its too big of a number though..

OpenStudy (anonymous):

is the unit meters?

OpenStudy (anonymous):

Yup

OpenStudy (anonymous):

remember that the ball is traveling at 360km/h or 223mph

OpenStudy (anonymous):

wait where did that number come form

OpenStudy (anonymous):

from

OpenStudy (anonymous):

I just converted from 100m/s to km/h

OpenStudy (anonymous):

dont worry about it if its confusing, just trust me that it's the right answer haha

OpenStudy (anonymous):

oh ok

OpenStudy (anonymous):

lol okay ill trust you

OpenStudy (anonymous):

ok, final question! no calculations needed for this one, how fast do you think it will be traveling when it hits the ground?

OpenStudy (anonymous):

i have no idea

OpenStudy (anonymous):

well we know the horizontal velocity stays the same, because no forces act on the ball in the horizontal direction. for the vertical we know the ball only every experiences one force, gravity, which is the same for when the ball is going up, and coming back down. so the ball will slow down, stop at the top, then start accelerating back to earth and be at tis initial velocity again once it reaches the ground!

OpenStudy (anonymous):

soo we factor in -9.81 to vy?

OpenStudy (anonymous):

yeah, but its the same going up and down, so the final velociy is the same as the initial velocity, 100 m/s

OpenStudy (anonymous):

ok so how would the equation look

OpenStudy (anonymous):

complicated.

OpenStudy (anonymous):

so what do i do

OpenStudy (anonymous):

your answer for c is 100m/s

OpenStudy (anonymous):

ok haha

OpenStudy (anonymous):

if asked to explain it by your teacher, just say you thought about conservation of energy

OpenStudy (anonymous):

ek = 1/2mv^2

OpenStudy (anonymous):

alright haah i have 2 more questions could you help ?

OpenStudy (anonymous):

how old are you?

OpenStudy (anonymous):

if you dont mind me asking

OpenStudy (anonymous):

17

OpenStudy (anonymous):

why. how about you

OpenStudy (anonymous):

21

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