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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
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) ]
Astoneisthrownverticallyupwardsfromthetopofatower50.0mhighwithaninitialvelocityof20.0ms1.calculatei.themaximumheightthestonereachesii.thetimeittakestoreachthemaximumheightiii.thetotaldistanceitcovers[takeg=10ms1]
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
i)andii)laalturaseralaqueyatienemaslaquegane.laalturafinalesdelaforma:y=yo+votaire12gtaire2=50+20×20.5×10×4=70mlavelocidadfinalvenelpuntomasaltoescero,eltiempoenelairetairees:v=vogtairetaire=vog=2010=2siii)LadistanciatotalYes:Y=yup+ydown=20+70=90m
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=20+70=90m
totaldistancecovered=20+70=90m
Commented by estudiante last updated on 29/Jan/19
Ooops :) You're right. I will change it. Thanks

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