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-maxi

Question Number 54015 by shaddie last updated on 28/Jan/19

$$\mathrm{A}\:\mathrm{stone}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{vertically}\:\mathrm{upwards}\:\mathrm{from}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{a}\:\mathrm{tower}\:\mathrm{50}.\mathrm{0m}\:\mathrm{high}\:\mathrm{with}\:\mathrm{an}\:\mathrm{initial}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{20}.\mathrm{0ms}^{−\mathrm{1}} .\mathrm{calculate} \\ $$$$\mathrm{i}.\mathrm{the}\:\mathrm{maximum}\:\mathrm{height}\:\mathrm{the}\:\mathrm{stone}\:\mathrm{reaches} \\ $$$$\mathrm{ii}.\mathrm{the}\:\mathrm{time}\:\mathrm{it}\:\mathrm{takes}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{height} \\ $$$$\mathrm{iii}.\mathrm{the}\:\mathrm{total}\:\mathrm{distance}\:\mathrm{it}\:\mathrm{covers}\left[\mathrm{take}\:\mathrm{g}=\mathrm{10ms}^{−\mathrm{1}} \right] \\ $$

Answered by estudiante last updated on 29/Jan/19

i) and ii) 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) −(1/2)gt_(aire) ^2 = 50+20×2−0.5×10×4=70m la velocidad final v en el punto mas alto es cero, el tiempo en el aire t_(aire) es: v=v_o −gt_(aire) → t_(aire) =(v_o /g)=((20)/(10))=2s iii) La distancia total Y es: Y=y_(up) +y_(down) =20+70= 90m

$$\left.{i}\left.\right)\:{and}\:{ii}\right) \\ $$$${la}\:{altura}\:{sera}\:{la}\:{que}\:{ya}\:{tiene}\:{mas}\:{la}\:{que}\:{gane}.\:{la}\:{altura}\:{final}\:{es}\:{de}\:{la}\:{forma}: \\ $$$$\:{y}={y}_{{o}} +{v}_{{o}} {t}_{{aire}} −\frac{\mathrm{1}}{\mathrm{2}}{gt}_{{aire}} ^{\mathrm{2}} =\:\mathrm{50}+\mathrm{20}×\mathrm{2}−\mathrm{0}.\mathrm{5}×\mathrm{10}×\mathrm{4}=\mathrm{70}{m} \\ $$$${la}\:{velocidad}\:{final}\:{v}\:{en}\:{el}\:{punto}\:{mas}\:{alto}\:{es}\:{cero},\:{el}\:{tiempo}\:{en}\:{el}\:{aire}\:{t}_{{aire}} \:{es}: \\ $$$${v}={v}_{{o}} −{gt}_{{aire}} \:\rightarrow\:{t}_{{aire}} =\frac{{v}_{{o}} }{{g}}=\frac{\mathrm{20}}{\mathrm{10}}=\mathrm{2}{s}\: \\ $$$$\left.{iii}\right) \\ $$$${La}\:{distancia}\:{total}\:{Y}\:\:{es}: \\ $$$${Y}={y}_{{up}} +{y}_{{down}} =\mathrm{20}+\mathrm{70}=\:\mathrm{90}{m}\: \\ $$

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

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