[tex]\huge\fbox\pink{@Siddhi163}[/tex]. → A ball at rest is dropped from a height of 12 m.If it looses 25% of its kinetic energy on striking the ground and bounces back to height 'h'is equal to?
The kinetic energy on reaching the ground will be equal to the initial potential energy, i.e. KE=mg(12).Since the remaining kinetic energy (after loss) will be used to rise the block to height h,mg(12)(1−0.25)=mghHence, h=9 m.
Answers & Comments
The kinetic energy on reaching the ground will be equal to the initial potential energy, i.e. KE=mg(12).
Since the remaining kinetic energy (after loss) will be used to rise the block to height h,
mg(12)(1−0.25)=mgh
Hence, h=9 m
Answer:
The kinetic energy on reaching the ground will be equal to the initial potential energy, i.e. KE=mg(12).Since the remaining kinetic energy (after loss) will be used to rise the block to height h,mg(12)(1−0.25)=mghHence, h=9 m.
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