A particle is projected from the ground with a kinetic energy E at an angle of 60° with the horizontal. Its kinetic energy at the highest point of its motion will be-
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thw total kinetic energy of the body when projected is E, E=1/2mv^2. at the highest point, the velocity in y direction is zero. ao the only kinetic energy in play here would be the kinetic energy in x direction which has been constant since the body was projected at, Ehprizontal= 1/2 m (vcos60)^2.
Samaira11111:
can u plz provide me the steps
theres not any steps to it actually, the total Kinetic Energy at any point in a projectile is equal to the kinetic energy in the vertical direction + the kinetic energy in horizontal direction. which is basically E= 1/2m(vsintheta)^2 + 1/2m(vcostheta)^2.now since the velocity in veetical direction at the highest point is 0, the kinetic energy at the highest point remains E= 1/2 m(vcosthwta)^2
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The total kinetic energy of the body when projected is E,E=1/2m v^ at the nighest point, the velocity in direction is zero.
The only kinetic energy in play heard would be the kinetic energy in x direction which has been constants since the body was projected at,
Elprizontal =1/2m (vcos60)^2.
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