Question Number 54015 by shaddie last updated on 28/Jan/19
![A stone is thrown vertically upwards from the top of a tower 50.0m high with an initial velocity of 20.0ms^(−1) .calculate i.the maximum height the stone reaches ii.the time it takes to reach the maximum height iii.the total distance it covers[take g=10ms^(−1) ]](Q54015.png)
$$\mathrm{A}\:\mathrm{stone}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{vertically}\:\mathrm{upwards}\:\mathrm{from}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{a}\:\mathrm{tower}\:\mathrm{50}.\mathrm{0m}\:\mathrm{high}\:\mathrm{with}\:\mathrm{an}\:\mathrm{initial}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{20}.\mathrm{0ms}^{−\mathrm{1}} .\mathrm{calculate} \\ $$$$\mathrm{i}.\mathrm{the}\:\mathrm{maximum}\:\mathrm{height}\:\mathrm{the}\:\mathrm{stone}\:\mathrm{reaches} \\ $$$$\mathrm{ii}.\mathrm{the}\:\mathrm{time}\:\mathrm{it}\:\mathrm{takes}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{height} \\ $$$$\mathrm{iii}.\mathrm{the}\:\mathrm{total}\:\mathrm{distance}\:\mathrm{it}\:\mathrm{covers}\left[\mathrm{take}\:\mathrm{g}=\mathrm{10ms}^{−\mathrm{1}} \right] \\ $$
Answered by estudiante last updated on 29/Jan/19
$$\left.{i}\left.\right)\:{and}\:{ii}\right) \\ $$$${la}\:{altura}\:{sera}\:{la}\:{que}\:{ya}\:{tiene}\:{mas}\:{la}\:{que}\:{gane}.\:{la}\:{altura}\:{final}\:{es}\:{de}\:{la}\:{forma}: \\ $$$$\:{y}={y}_{{o}} +{v}_{{o}} {t}_{{aire}} −\frac{\mathrm{1}}{\mathrm{2}}{gt}_{{aire}} ^{\mathrm{2}} =\:\mathrm{50}+\mathrm{20}×\mathrm{2}−\mathrm{0}.\mathrm{5}×\mathrm{10}×\mathrm{4}=\mathrm{70}{m} \\ $$$${la}\:{velocidad}\:{final}\:{v}\:{en}\:{el}\:{punto}\:{mas}\:{alto}\:{es}\:{cero},\:{el}\:{tiempo}\:{en}\:{el}\:{aire}\:{t}_{{aire}} \:{es}: \\ $$$${v}={v}_{{o}} −{gt}_{{aire}} \:\rightarrow\:{t}_{{aire}} =\frac{{v}_{{o}} }{{g}}=\frac{\mathrm{20}}{\mathrm{10}}=\mathrm{2}{s}\: \\ $$$$\left.{iii}\right) \\ $$$${La}\:{distancia}\:{total}\:{Y}\:\:{es}: \\ $$$${Y}={y}_{{up}} +{y}_{{down}} =\mathrm{20}+\mathrm{70}=\:\mathrm{90}{m}\: \\ $$
Commented by estudiante last updated on 28/Jan/19
May I be right?
Commented by mr W last updated on 28/Jan/19
$${total}\:{distance}\:{covered}=\mathrm{20}+\mathrm{70}=\mathrm{90}{m} \\ $$
Commented by estudiante last updated on 29/Jan/19
Ooops :) You're right. I will change it. Thanks