Question

Could you please help me solve part b?
https://www.dropbox.com/s/3crr9qqfqplsr9d/Screenshot%202013-11-02%2007.24.44.jpg

Answer 1

@ash2326

Answer 2

@e.mccormick

Answer 3

@Hero

Answer 4

@Luigi0210

Answer 5

@timo86m

Answer 6

@zepdrix

Answer 7

Do you use kinematic equations?

Answer 8

I am afraid that this is my first time in this section and somethings that can be consider basic to others to me are unfamiliar

Answer 9

@Christos From the graph can you find the max height?

Answer 10

its more or less 200 right ?

Answer 11

'about' 200

Answer 12

I am going to watch how ash solves this problem.

Answer 13

Cool, can you find the corresponding horizontal distance?

Answer 14

from the graph ?

Answer 15

yes

Answer 16

about 6.5

Answer 17

okay, let's just look at the vertical motion, let the vertical component of velocity be u
Now the projectile is launched, Is there any force acting in vertical direction?

Answer 18

the diagonal ?

Answer 19

x-axis ?

Answer 20

Christos read the first line of the question

Answer 21

may it be gravity ?

Answer 22

I dont understand these 2 terms: trajectory and projectile

Answer 23

yes, it acts vertically downwards. So when our projectile is launched, it's being slowed every moment. Until its vertically at rest, that's the point of it's max height.

Answer 24

Please not it'll be at rest only for a moment, then it'll start coming down.

Answer 25

I see

Answer 26

Trajectory is the shape of the motion, suppose if a man moves straight in line. Its trajectory will be straight.

Answer 27

the velocity starts from 0 ?

Answer 28

Christos we can't have velocity start from 0, that would be the case of a rocket, it has a propulsion system which will accelerate it and increase its velocity from 0.
Here on the other hand, we impart a velocity to the projectile. For example if you throw a ball straight up, you impart velocity to the ball only in vertical direction. Suppose 50m/s, as it goes up, it'll start to slow down.

Answer 29

here the ball will go straight up and come down straight down. As we have not given it any vertical velocity, it'll come down straight in your hand. There will be no horizontal motion

Answer 30

oh

Answer 31

Are you followin?

Answer 32

I mean horizontal sorry.

Answer 33

so in this problem I just throw something up

Answer 34

Nope that's the catch, you can see it moving along in x direction as well. So you're imparting velocity in both horizontal and vertical direction

Answer 35

oh

Answer 36

Like the goal keeper kicks the ball in a football match. ball moves both horizontally and vertically

Answer 37

Do you follow till here.

Answer 38

Yes I understand

Answer 39

Cool, so for finding only the vertical velocity, just focus at vertical motion. Ball moves up, reach the highest point and comes back down.
Let the initial vertical velocity be u, it'll reach the highest point at that point vertical velocity would be zero, wouldn't it ?

Answer 40

yes thats right

Answer 41

0 and then it will start falling

Answer 42

yes, so do you know the max height?

Answer 43

ok so when v = 0 hight = 200

Answer 44

yes, use this
\[v^2-u^2=2as\]
v=0
s=200
a=gravity
u=initial vertical velocity

Answer 45

ok hold

Answer 46

Christos one important thing

Answer 47

sqrt(3920)

Answer 48

yes ?

Answer 49

the vertical initial velocity is upwards and gravity is downwards. Make sure to assign them opposite signs, if vertical direction upwards is +ve then
\[u=+ve\]
\[a=-g\]

Answer 50

will this change my result ?

Answer 51

like -sqrt(3920)

Answer 52

because I just can't take it inside lol

Answer 53

the minus

Answer 54

ohhh it will become + because its already -

Answer 55

if you assume upwards as +ve and downwards as -ve it'll be the same
\[u=\sqrt{3920}\]

Answer 56

so your part b is solved, do you understand?

Answer 57

ok got it

Answer 58

what type of formula do I need to use in order for me to solve
c ?

Answer 59

That can be easily solved, as we have the initial vertical velocity now.
\[v-u=at\]
v=0.
u= you just found out
a=-g
t=????

Answer 60

ok its my previews answer divided by gravity and a minus

Answer 61

Time , it has to be positive. Remember that there is a -ve sign with g
yes, other wise it is correct

Answer 62

oh so its positive because its time

Answer 63

yes,

Answer 64

for d I have to use V=dx4/dt ?

Answer 65

Nope, this is also straight forward. When you launched the projectile, you gave it a horizontal velocity h
Now you already know the time for max height, at that time from the graph you can figure out the horizontal distance.
Divide distance by time to get the horizontal velocity.

Answer 66

horizontal distance is 6.5 ??

Answer 67

yes,

Answer 68

oh so that would be 63.7/sqrt(3920)

Answer 69

now there has to be some angle formula I think

Answer 70

You used the time for max height and horizontal distance to find the horizontal velocity, didn't you?

Answer 71

yes

Answer 72

good, it's easy to find angle.
\[\tan \theta=\frac{vertical}{horizontal}\]