(a) Find the initial speed.
Solution
The relevant equations are (4) and (5) derived here and reproduced below as (1) and (2) respectively:
$$\begin{align}
& \frac{\Delta x}{\tan\!\theta-\tan\!\varphi}=\frac{2v_{0x}^2}{g} \\
& \tan\!\varphi=\frac{1}{2}(\tan\!\theta+\tan\!\omega)...
U1 = -GMm/r
K1 = (1/2)mvi^2
U2 = as r approaches infinity, U2 approaches zero
K2 = (1/2)mvf^2
(1/2)mvi^2 - GMm/r = (1/2)mvf^2 + 0
vi = √(vf^2 + (2GMm)/r) = √(250,000 + 2(6.7 E-11)(6 E23)/3400) = 153776.815
But that is not the correct answer, can anybody see my mistake/misunderstanding?
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