Question

The position of projectile launched from the origin at t = 0 is given by $\vec{r} = (40 \hat{i} + 50 \hat{j}) m$ at t = 2s. If the projectile was launched at an angle $θ$ from the horizontal, then $θ$ is (take g = 10 m/s²).

Moderate

Solution

The position vector of the projectile is $\vec{r} = \vec{ut} + \frac{1}{2} {\vec{at}}^{2}$
$\begin{array}{l} 40 \hat{i} + 50 \hat{j} = \vec{u} \times 2 + \frac{1}{2} (- 10 \hat{j}) \times 2^{2} \\ \Rightarrow \vec{u} = u_{x} \hat{i} + u_{y} \hat{j} = 20 \hat{i} + 35 \hat{j} \\ \tan θ = \frac{u_{y}}{u_{x}} = \frac{35}{20} = \frac{7}{4} \Rightarrow θ = \tan^{- 1} (\frac{7}{4}) \end{array}$

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