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$${ v }_{ 0 }=\dfrac { x }{ cos{ \theta }_{ 0 } } \sqrt { \dfrac { g }{ 2(xtan{ \theta }_{ 0 }-y) } } $$

$${ v }_{ 0 }= 255.5 ≈ 2.6 × { 10 }^{ 2 }m/s$$

(b) For projectile motion $$t$$ is given by formula

$$t=\dfrac { x }{ { v }_{ 0 }{ cos\theta }_{ 0 } } $$

(c) We expect the air to provide resistance but no appreciable lift to the rock, so we would need a greater launching speed to reach the same target.

putting $$x = 9400 m$$, $$y = –3300 m$$, and $${ \theta }_{ 0 } = 35°$$.

$${ v }_{ 0 }=\dfrac { 9400 }{ \cos35^o } \sqrt { \dfrac { g }{ 2(x\tan35^o-3300) } } $$

$${ v }_{ 0 }= 255.5 ≈ 2.6 × { 10 }^{ 2 }m/s$$

(b) For projectile motion $$t$$ is given by formula

$$t=\dfrac { x }{ { v }_{ 0 }{ cos\theta }_{ 0 } } $$

$$t=\dfrac { 9400m }{ (255.5m/s)cos{ 35 }^{ o } } =45s.$$

(c) We expect the air to provide resistance but no appreciable lift to the rock, so we would need a greater launching speed to reach the same target.

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