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Q.

A Projectile travels at an angle $θ$ below horizontal, 1sec after projection from origin as shown. What is the value of $θ$ ? (Take g= $10 m / s^{2}$ )

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a

$\tan^{- 1} (\frac{1}{2})$

b

$\tan^{- 1} (2)$

c

$37 °$

d

$53 °$

answer is B.

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Detailed Solution

$\begin{array}{l} \tan θ = \frac{v_{y}}{v_{x}} = \frac{u_{y} - g t}{u_{x}} u_{x} = \frac{4}{1} m / s 3 = u_{y} \times 1 - \frac{1}{2} \times 10 \times 1^{2} u_{y} = 8 m / s \\ \Rightarrow \tan θ = | \frac{8 - 10}{4} | = | - \frac{1}{2} | \end{array}$

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A Projectile travels at an angle θ below horizontal, 1sec after projection from origin as shown. What is the value of θ ? (Take g= 10 m/s2)