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Question Number 190266 by alcohol last updated on 30/Mar/23
a ball is thrown vertically upward  from a point 0.5m above the ground with  speed u = 7m/s  find the height reached above ground  g = 10m/s^2
$${a}\:{ball}\:{is}\:{thrown}\:{vertically}\:{upward} \\ $$$${from}\:{a}\:{point}\:\mathrm{0}.\mathrm{5}{m}\:{above}\:{the}\:{ground}\:{with} \\ $$$${speed}\:{u}\:=\:\mathrm{7}{m}/{s} \\ $$$${find}\:{the}\:{height}\:{reached}\:{above}\:{ground} \\ $$$${g}\:=\:\mathrm{10}{m}/{s}^{\mathrm{2}} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 30/Mar/23
v^2 =u^2 −2gh   0=7^2 −2(10)h ⇒h=((49)/(20)) m  height=h+0.5=2.95 m
$$\mathrm{v}^{\mathrm{2}} ={u}^{\mathrm{2}} −\mathrm{2}{gh}\: \\ $$$$\mathrm{0}=\mathrm{7}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{10}\right){h}\:\Rightarrow{h}=\frac{\mathrm{49}}{\mathrm{20}}\:{m} \\ $$$$\mathrm{height}={h}+\mathrm{0}.\mathrm{5}=\mathrm{2}.\mathrm{95}\:{m} \\ $$

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