At time t = 0, a projectile is launched from ground level. At t = 2.00 s, it is displaced d = 43 m horizontally and h = 59 m vertically above the launch point. What are the (a) horizontal and (b) vertical components of the initial velocity of the projectile? (c) At the instant it reaches its maximum height above ground level, what is its horizontal displacement D from the launch point?

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Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.