A-tennis-ball-is-thrown-vertically-upward-with-an-initial-velocity-of-50m-s-when-the-ball-return-to-the-point-of-projection-it-renounce-with-the-velocity-of-2-3-of-the-velocity-Calculate-the-hei

Question Number 10736 by Saham last updated on 23/Feb/17

$$\mathrm{A}\:\mathrm{tennis}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{vertically}\:\mathrm{upward}\:\mathrm{with}\:\mathrm{an}\:\mathrm{initial}\:\mathrm{velocity}\:\mathrm{of} \\ $$$$\mathrm{50m}/\mathrm{s},\:\mathrm{when}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{return}\:\mathrm{to}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{projection}\:\mathrm{it}\:\mathrm{renounce}\: \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{of}\:\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{of}\:\mathrm{the}\:\mathrm{velocity}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{heigth}\:\mathrm{after}\:\mathrm{renounce}. \\ $$

Answered by mrW1 last updated on 24/Feb/17

(1/2)mV_2 ^2 =mgh ⇒h=(V_2 ^2 /(2g))=((((2/3)×50)^2 )/(2×9.81))=56.6 m

$$\frac{\mathrm{1}}{\mathrm{2}}{mV}_{\mathrm{2}} ^{\mathrm{2}} ={mgh} \\ $$$$\Rightarrow{h}=\frac{{V}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}{g}}=\frac{\left(\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{50}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{9}.\mathrm{81}}=\mathrm{56}.\mathrm{6}\:{m} \\ $$

Commented by Saham last updated on 24/Feb/17