Question

A projectile of mass 0.50 Kg is fired with an initial speed of 10 m.s at an angle of \(60^o\) above the horizontal. The potential energy (relative to the ground level) of the projectile at its highest point is...?

Answer 1

do i use \[U_{GRAV} = mgy\]

Answer 2

Direction : put vertical component of speed to 0 leave horizontal as it is , This the speed at the Apex of the parabola. Calculate the absolute reduction of the E-kinetic this is your potential energy

Answer 3

Calculate maximum height reached..
Vertical velocity = 10 sin 60
A = -10
calculate from v^2-u^2 = 2as..
multiply height by M*g

Answer 4

NO NEED to calculate the height directly -ust as I have told you use the reduction i E-kin

Answer 5

final velocity is zero
initial is 10 sin60

Answer 6

lol.. whatever :P

Answer 7

wait..apex parabola? what's that?

Answer 8

and absolute reduction?

Answer 9

lmfau - You probably don't realize that one does not need to know the top height to know the max potential enrgy, do ya me LOL..

Answer 10

so much fancy terms

Answer 11

My remark addresse to vikrang

Answer 12

Again lgb listen here:

Answer 13

you can do it either way :S

Answer 14

Calculate the initial Vertical velocity

Answer 15

Then lgb, calculate the m* Vvertical^2 /2

Answer 16

initial vertical velocity is 0 right?

Answer 17

oh wait...that's final vertical velocity

Answer 18

Thsi is Your solution for maximal potential energy

Answer 19

at max height v = 0 right?

Answer 20

\[V_{y, init} = V*Sin \alpha \]

Answer 21

And of course at Top V_y = 0 (NOT V though - the body keeps flying)

Answer 22

is that initial V_y = V sin \(\alpha\)

Answer 23

Yes

Answer 24

so how do i know what V is?

Answer 25

oh 10 m/s

Answer 26

yes. that makes sense

Answer 27

\[V_{apex}\neq0\]

Answer 28

\[V_{y_i} = 10 \sin (60) \implies 5\sqrt 3\]
right?

Answer 29

okay...what's V_apex now?

Answer 30

The total speed at the top (Apex is Top but latex has Top as a command lol...)

Answer 31

why isnt the v at top 0?

Answer 32

It should have been simple if you have had calculated height :/

Answer 33

THIS IS SOMETHING CENTRAL YOU MUST READ IN BOOKS. SHORT ANSWER IS:
V-Horizontal is CONSTANT during the whole process. Of course ubder the idealization of no-air-friction

Answer 34

SO LGB : V_Horizontal is CONSTANT, and in your data it is NONZERO - so V_total is non-zero at the top

Answer 35

lol i agree @vikrantg4 =)) these shortcuts make me confuse

Answer 36

I repeat\[V_{x} = CONSTANT\]

Answer 37

but at max height it will stop... so velocity there should be 0

Answer 38

and at the top \[V = V_{x}\]

Answer 39

Nah.. It stops only going up.. It continues to travel in Horizontal direction

Answer 40

horizontal?

Answer 41

but it's fired at 60 degree angle

Answer 42

This shows you the projectile DOES NOT STOP AT THE TOP - IT FLIES at that instant of course - HORIZONTALLY

Answer 43

omg.. at TOP it stops travelling vertically and continues in horizontal direction

Answer 44

You know resolution of vectors ? eh .. :p

Answer 45

no?

Answer 46

Lgb : The HOrizontal component of velocity\[V_{x}\] stays OCNSTANT and it is NOT ZERO !

Answer 47

i know the horizontal velocity is constant..what im asking is V_y

Answer 48

V_y should be 0 at max height

Answer 49

yes

Answer 50

Your statement "the projectile stops" is wrong.

Answer 51

that's what i was aksing...

Answer 52

V_y at grtound is 10 sin60

Answer 53

just put conservation of energy and calculate

Answer 54

It keeps flying with \[V_{x}\]

Answer 55

....so this doesnt use mgy?

Answer 56

You have TWO methods available
A. As I solved for you - just put the difference in kinetic energies and you have the answer
B. What Vikrang4 suggets - find the top height and , yes, plug it into m*g*y

Answer 57

See my detailed suggestion at the beginning of the exchange

Answer 58

to get V i have to do \(\sqrt{V_x^2 + V_y^2}\) first right?

Answer 59

NOOO

Answer 60

no?

Answer 61

oh wait..why did i do that

Answer 62

At the top only\[V_{x} \] survuves

Answer 63

nevermind what i said lol

Answer 64

The \[V_{y} = 0\] at the top point

Answer 65

one more question... \[\huge \Delta U = \Delta K\]??

Answer 66

Yes but with a small correction - a MINUS sign