First slide

Projection Under uniform Acceleration

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).

Easy

A

$\tan^{- 1} (2 / 3)$
B

$\tan^{- 1} (3 / 2)$
C

$\tan^{- 1} (7 / 4)$
D

$\tan^{- 1} (4 / 5)$

Solution

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

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