A ball is dropped from rest at a height of 60 m . On striking the ground, it loses 25% of its of its energy. To what height does it rebound?

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Solution

R.E.F image
$V^{2} = V^{2} + 2 a s$
$V = \sqrt{2 \times 10 \times 60} = \sqrt{1200} m / s$
KE initial $= \frac{1}{2} m v^{2}$
$= 600 m$
KE initial $= 0.75$ KE initial
$= 450 m$
$\frac{1}{2} m v^{12} = 450 m$
$V^{1^{2}} = 900$
$V^{112} = V^{12} + 2 a h$
$O = 900 - 2 \times 10 \times h$
$h = 45 m$

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A ball is dropped from rest at a height of 60 m . On striking the ground, it loses 25% of its of its energy. To what height does it rebound?

R.E.F imageV2=V2+2asV=√2×10×60=√1200m/sKE initial =12mv2=600mKE initial =0.75 KE initial =450m12mv12=450mV12=900V112=V12+2ahO=900−2×10×h h=45m

R.E.F image
$V^{2} = V^{2} + 2 a s$
$V = \sqrt{2 \times 10 \times 60} = \sqrt{1200} m / s$
KE initial $= \frac{1}{2} m v^{2}$
$= 600 m$
KE initial $= 0.75$ KE initial
$= 450 m$
$\frac{1}{2} m v^{12} = 450 m$
$V^{1^{2}} = 900$
$V^{112} = V^{12} + 2 a h$
$O = 900 - 2 \times 10 \times h$
$h = 45 m$