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Question Number 11536 by tawa last updated on 27/Mar/17

Answered by ridwan balatif last updated on 28/Mar/17

$$\mathrm{Data}:\mathrm{m}=0.140\mathrm{kg},v_{\mathrm{o}}=25\mathrm{m}/\mathrm{s}$$$$\left(\mathrm{a}\right)\mathrm{EK}=\frac{1}{2}\mathrm{m}{v}_{\mathrm{o}}^2$$$$=\frac{1}{2}\times 0.140\times {25}^2$$$$=43.75\mathrm{J}$$$$\left(\mathrm{b}\right)\mathrm{if}\mathrm{the}\mathrm{ball}\mathrm{wanna}\mathrm{reach}\mathrm{maximum}\mathrm{height}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{velocity}\mathrm{at}\mathrm{maximum}\mathrm{height}\mathrm{must}\mathrm{be}\mathrm{zero}(v_{\mathrm{t}}=0)$$$$v_t^2=v_{\mathrm{o}}^2-2gh$$$$0^2={25}^2-2\times 10\times h$$$$h=32.5m$$$$\left(\mathrm{c}\right)v^2=v_o^2-2g\mathrm{h}$$$$v^2={25}^2-2\times 10\times 30$$$$v=5\mathrm{m}/\mathrm{s}$$$$\mathrm{EK}=\frac{1}{2}\mathrm{m}{v}^2$$$$=\frac{1}{2}\times 0.14\times 5^2$$$$=1.75\mathrm{J}$$$$$$

Commented by tawa last updated on 28/Mar/17

$$\mathrm{I}\mathrm{really}\mathrm{appreciate}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

Answered by Nysiroke last updated on 28/Mar/17

$$whereh=?$$$$v^2={\mathrm{u}}^2-2\mathrm{gh}(\mathrm{upward}\mathrm{is}-\mathrm{ve})$$$$\mathrm{given}\mathrm{that}\mathrm{u}={25\mathrm{ms}}^{-1},v=0,g=10{ms}^{-2}$$$$so$$$$0^2={25}^2-2\times 10\times h$$$$0=625-20h$$$$\because 20h=625$$$$h=\frac{625}{20}$$$$=125/4=31.25m(max.heightreachedbytheball)$$$$$$$$a)totalenergyoftheball=K.E+mgh$$$$=\frac{1}{2}{mu}^2+mgh$$$$=\frac{m}{2}(u^2+gh),whereh=31.25m$$$$=\frac{0.140}{2}({25}^2+10\times 31.25)$$$$=0.070(625+312.5)$$$$=0.070(937.5)$$$$=65.63J$$$$ci)velocityat30mupwardisgivenby$$$$v^2=2gh$$$$v^2=2gh$$$$=2\times 10\times 30=600$$$$\therefore v=\sqrt{600}$$$$=24.50{ms}^{-1}$$$$cii)k.eat30misgivenby$$$$KE=\frac{1}{2}{mv}^2$$$$=\frac{1}{2}\times 0.140\times \sqrt{{600}^2}$$$$=0.070\times 600$$$$=42J$$$$$$$$$$

Commented by tawa last updated on 28/Mar/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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