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

$\vec{r}= \left ( 40\hat{i} +50\hat{j}\right )m$ at t = 2s.

If the projectile was launched at an angle $\Theta$ from the horizontal, then $\Theta$ is (take g=10 ms^-2).

Option: 1
$tan ^{-1}\: 2/3$

Option: 2
$tan ^{-1}\: 3/2$

Option: 3
$tan ^{-1}\: 7/4$

Option: 4
$tan ^{-1}\: 4/5$

Answers (1)

best_answer

Using

$S=ut+\frac{1}{2}at^2$

$u Cos\theta\times t+ \frac{1}{2}(0)t^{2}=\:u t\:\:\:\:\:\:\:\:\:\ \because g=0$

$u Cos\theta.t=40$

$(u Cos\theta) (2)=40$

$u Cos\theta = 20 -------- (1)$

$u Sin\theta t - \frac{1}{2}\:gt^{2}=50$

$(u Sin\theta \times 2)-(\frac{1}{2}\times10\times4)=50$

$u Sin\theta=35----------(2)$

From equation (1) and (2)

$\theta = tan^{-1}(\frac{7}{4})$

Posted by

shivangi.bhatnagar

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