1.

If K is the kinetic energy of a projectile fired at an angle 45°, then what is the kinetic energy at the highest point.

Answer»

`(K)/(4)`
`(K)/(2)`
K
2K

SOLUTION :`k=(1)/(2)mv^(2)`
At highest pt., `NU = v cos45^(@)`
` K.E =(1)/(2) m((v)/(SQRT(2)))^(2)=(v)/(sqrt(2))`
`=(1)/(2) (mv^(2))/(2)=(K)/(2)`


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