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First resolve the vertical component of your equation with the formula: Fsinθ (In layman's terms, Force X Sine X anglethrown) This is your initial velocity. At maximum height, velocity is ALWAYS 0. So: Since: V^2 = U^2 + 2as Then: S=(V^2-U^2) / 2a Eg: If we threw a ball at 26ms with an angle of 64: 26sin64=23.36 S= We want to find this value U=23.36 V=0 A=9.81 T= We don't need to know this. S= (0^2 - 545.68^2) / 2x-9.81 S= -545.68 / -19.62 S= 27.8m So the maximum height would be 27.8m Remember, at maximum vertical height V is ALWAYS 0!!!

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14y ago

Kinetic Energy or Potential energy= mass x gravity x height

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Q: How do you find the maximum height in a physics question?
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