The position of a projectile launched from the origin at t = 0 is given by → r = (4 0 ^ i + 5 0 ^ j ) m at t = 2s. If the projectile was launched at an angle θ from the horizontal, then θ is (take g = 10 ms -2 ).

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Correct answer: C. ^ - 1 (7)/(4)

The position of a projectile launched from the origin at t = 0 is given by → r = (4 0 ^ i + 5 0 ^ j ) m at t = 2s. If

From question, Horizontal velocity (initial),
${u}_{x} = \dfrac{40}{2} = 20m / s$
Vertical velocity (initial), $50 = {u}_{y}t + \dfrac{1}{2}g{t}^{2}$

$⇒ 50 = {u}_{y} \times 2 + \dfrac{1}{2} ( - 10 ) \times 4$
$50 = 2{u}_{y} - 20$
${u}_{y} = \dfrac{70}{2} = 35m / s$
$∴ tan\theta = \dfrac{{u}_{y}}{{u}_{x}} = \dfrac{35}{20} = \dfrac{7}{4}$
$⇒$ $\theta = \tan^{- 1}\dfrac{7}{4}$

The position of a projectile launched from the origin at t = 0 is...

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