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OpenStudy (anonymous):

a projectile of mass 50 gm is fired with velocity 10m s at an angle 30 with the horizontal calculate the linear momentumand kinetic energy of the projectile at highest point?

OpenStudy (ash2326):

At the highest point the vertical component of the velocity will be zero, do you agree with this?

OpenStudy (anonymous):

becoz it was fired horizantly

OpenStudy (ash2326):

|dw:1341756910842:dw| If it was given only vertical velocity, then also at the highest point it'd be zero

OpenStudy (anonymous):

at the highest point linear momentum would be mucos theta

OpenStudy (anonymous):

\[v ^{2}= u ^{2} - 2as\] u is 10 sin theta, a is acc due to gravity, find s, then we use it to find potential energy which is mgs which we get 125 so energy is 125

OpenStudy (anonymous):

can we do lik this? k.e= 1/2 m sq(u cos theta)

OpenStudy (anonymous):

sorry it is 6.25, @mathavraj i thought it is safe to just see change in height

OpenStudy (anonymous):

@mathavraj i think we can. ucos theta is the velocity at the highest point. sub it in formula for KE should give us the answer.

OpenStudy (anonymous):

i got KE as 1.87 J

OpenStudy (anonymous):

@Ruchi what is the answer?

OpenStudy (anonymous):

@Vaidehi09 what did u do?

OpenStudy (anonymous):

KE = 1/2 m u^2 cos^2(30)

OpenStudy (anonymous):

it should be sin thta na?

OpenStudy (anonymous):

theta is the angle with the horizontal. so horizontal component of velocity will have cos..not sin

OpenStudy (anonymous):

no u sin theta is zero at top

OpenStudy (anonymous):

yes and ke energy is chaange in velocity so it will be 1/2m(vi-vf)

OpenStudy (anonymous):

pls explain @rdx

OpenStudy (anonymous):

well kinetic energy is change in its intial energy and final energy so taking sin theta or cos theta should not make a difernece ,

OpenStudy (anonymous):

so let me do this by taking cos component in consideration \[1/2m(v \cos 30-v \cos 0)^{2}\]

OpenStudy (anonymous):

check it wid sin shoot me if u get any other result :P

OpenStudy (anonymous):

@mathavraj any problem??

OpenStudy (anonymous):

its zero wen we use sin component..but i still dint get that change thing..at the highst point there is only cos component...wich is constant throught the motion

OpenStudy (anonymous):

umm wait

OpenStudy (anonymous):

sorry i made a mistake again but 6.25 is correct. u urself agree that that there is no change in cos componet so there is no change in kinetic energy in x direction but only change in kinetic energy y direction and initially K.E in y direction is 6.25 and finally it is 0 so initial - fianl is the net K.E

OpenStudy (anonymous):

so should we take sin also

OpenStudy (anonymous):

what do u wnat to say?

OpenStudy (anonymous):

@Ruchi. what is the answer?

OpenStudy (anonymous):

i don't know its answer is nt mention

OpenStudy (anonymous):

we don;t hav to use its mass

OpenStudy (anonymous):

?

OpenStudy (anonymous):

y not mass?

OpenStudy (anonymous):

1/2"M"v^2

OpenStudy (anonymous):

@Rdx ?

OpenStudy (anonymous):

yes what do u want? u will need mass to find the energy, whenever a qustion asks potential or kinetic energy u actually need to calculate change in potential or kinetic energy

OpenStudy (anonymous):

best is solve by seeing change in height

OpenStudy (anonymous):

@Rdx KE at the highest point = KE in x direction + KE in y direction.....right? but velocity in y direc at that pt = 0 so KE at highest pt = KE in x direc........or am i wrong?

OpenStudy (anonymous):

IT IS CHANGE IN KINETIC ENERGY,

OpenStudy (anonymous):

well the x componet of velocity remains constant throughot and the change is y componet |dw:1341827827130:dw|

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