Question

a ball is thrown vertically up with a velocity 20 m/s. at what height will it's kinetic energy be 1/2 of its original value

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Answer 1

by law of conservation of energy..

$\frac{1}{2} m {v}^{2} = mgh$

$\frac{ \frac{1}{2} \times \frac{1}{2} m {v}^{2} }{mg} = height$

$\frac{ {20}^{2} }{4 \times 10} = h$

$\frac{400}{40} = 10 = h$

so at height 10metre the value kinetic energy will be half...