# A ball of mass 0/2 jg is allowed to fall from a height of 5n and after hitting thr floor bounces back to a height of 4m determine the impulse the ball is subjected to if the time of impact is 0.1 sec estimate the force experienced by the ball

The question is not clear enough as mass is not clearly defined. I assume it to be 0.2 kg.

The ball is dropped from a height 5 m so its velocity before it hit the floor is given by

${v}^{2}=0+2\times 10\times 5\phantom{\rule{0ex}{0ex}}v=10m/s\phantom{\rule{0ex}{0ex}}soitsmomentumwillbe{p}_{1}=-mv=-0.2\times 10=-2kg\xb7m/s\phantom{\rule{0ex}{0ex}}-ivesignisbecausethevelocityofballisdownward.\phantom{\rule{0ex}{0ex}}afterithitthefloorandbouncesbacktotheheight4msothevelocityof\phantom{\rule{0ex}{0ex}}balljustafterithitthefloorisgivenby\phantom{\rule{0ex}{0ex}}0={v}^{2}-2\times 10\times 4\phantom{\rule{0ex}{0ex}}v=4\sqrt{5}m/s=8.94m/s\phantom{\rule{0ex}{0ex}}andsomomentumis{p}_{2}=0.2\times 8.94=1.78kg\xb7m/s\phantom{\rule{0ex}{0ex}}sochangeinmomentumis\phantom{\rule{0ex}{0ex}}{p}_{2}-{p}_{1}=1.78-\left(-2\right)=3.78kgm/s\phantom{\rule{0ex}{0ex}}butimpulseisgivenbyI=F\xb7\u2206t={p}_{2}-{p}_{1}\phantom{\rule{0ex}{0ex}}F=\frac{3.78}{0.1}=37.8N$

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