Question

A ball is thrown vertically up with a velocity 20 m/s.At what height will it's kinetic energy be half of it's original kinetic energy?
please help me fast​

Answer 1

Answer:

Initial velocity u = 20 m/s

Let rhe mass me m kg.

Initial K.E= ½mu²= ½m20²= 200m J.

Let the velocity at the point where its K.E becomes half its original value be v.

K.E at that point = ½mv²

K.E at that point is equal to half of the initial K.E.

½mv²= ½200m

=> mv²= 200m

=>v²= 200

=> v= 10√2 m/s.

u = 20 m/s , v = 10√2 m/s , g= -10 m/s²

v²- u²= 2gh

=> 10√2²- 20²= 2*-10*h

= > -200= -20h

=> h = 10m.

Hence at height of 10m , its K.E will be equal to half its original value.

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