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anonymous

  • 4 years ago

An object is thrown vertically upward and has a speed of 10 m/s when it reaches three fifths of its maximum height above the launch point. Determine its maximum height.

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  1. anonymous
    • 4 years ago
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    i used the 2ad = V(inital)^2 - V^2 formula H = (1/5), a = -9.8, V^2 = 10m/s since intial V is 0 so i had 2(-9.8)d = 0^2 - 10^2 -19.6d = -100 => d= -100/-19.6 => d=5.10m after this i multiplied it by the reciprocal of 1/5 and got 25.5m i'm stuck on where i went wrong?

  2. ash2326
    • 4 years ago
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    We know \[v^2-u^2=2as\] v= final velocity at max height is 0 u = u ( initial velocity) s= h (max height ) a=-9.8 m/s^2 so we get \[-u^2=-2*9.8*h.............equation 1\] now at s= 3/5 h v= 10 m/s u= u m/s so from this we get \[10^2-u^2=-2*9.8*\frac{3}{5}*h..........equation 2\] subtract equation 1 from 2 we get \[10^2=-2*9.8*\frac{3}{5}h+2*9.8*h\] on simplifying we get \[100=2*9.8*\frac{1}{5}*h\] so \[h=\frac{100*5}{2*9.8}\] \[h = 25.51 m\] so max height is 25.51 meters

  3. ash2326
    • 4 years ago
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    you are correct galactic

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