Question Number 86479 by Rio Michael last updated on 28/Mar/20
$$\mathrm{a}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{droped}\:\mathrm{from}\:\mathrm{a}\:\mathrm{height}\:\mathrm{20}\:\mathrm{m}.\:\mathrm{Given}\:\mathrm{that}\:\mathrm{it} \\ $$$$\mathrm{rebounce}\:\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{of}\:\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{that}\:\mathrm{which}\:\mathrm{it}\:\mathrm{hit}\:\mathrm{the}\:\mathrm{ground} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{time}\:\mathrm{interval}\:\mathrm{between}\:\mathrm{the}\:\mathrm{first}\:\mathrm{and}\:\mathrm{second}\:\mathrm{rebounce}. \\ $$
Answered by mr W last updated on 29/Mar/20
$${the}\:{height}\:{after}\:{a}\:{rebounce}\:{is}\:\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \:{of} \\ $$$${the}\:{height}\:{before}\:{the}\:{rebounce}. \\ $$$${h}_{\mathrm{0}} =\mathrm{20}\:{m} \\ $$$${h}_{\mathrm{1}} =\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \mathrm{20}=\mathrm{11}.\mathrm{25}\:{m} \\ $$$${between}\:{the}\:{first}\:{and}\:{second}\:{rebounce} \\ $$$${the}\:{ball}\:{reaches}\:{a}\:{height}\:{of}\:\mathrm{11}.\mathrm{25}{m} \\ $$$${and}\:{falls}\:{down}\:{again}.\:{the}\:{time}\:{needed} \\ $$$${is}\:{t}_{\mathrm{1}} =\mathrm{2}\sqrt{\frac{\mathrm{2}{h}_{\mathrm{1}} }{{g}}}=\mathrm{2}\sqrt{\frac{\mathrm{2}×\mathrm{11}.\mathrm{25}}{\mathrm{10}}}=\mathrm{3}\:{sec} \\ $$
Commented by Rio Michael last updated on 29/Mar/20
$${thanks}\:{sir} \\ $$