The position of a projectile launched from the origin at $t=0$ is given by $\overrightarrow{r}=(40\hat{i}+50\hat{j})\text{m}$ at $t=2\text{s}$. If the projectile was launched at an angle $\theta$ from the horizontal, then $\theta$ is
$(\text{Take}g=10{\text{ms}}^{-2})$

Q. The position of a projectile launched from the origin at $t=o$ is given by $\overrightarrow{r}$ =$(40\hat{i}+50\hat{i})$ m at $t=2s$. If the projectile was launched at an angle $\theta$ is (Take $g=10m{s}^{-2}).$

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

Q. A projectile is launched with a speed of 10 m/s at an angle ${60}^{\circ }$ with the horizontal from a sloping surface of inclination ${30}^{\circ }$. The range R is $(Takeg=10m/s^2)$

Q. The range in meters of a projectile launched over a flat ground from the origin with positive velocity $V$ in $m/s$ at an angle $\theta$ given in radian is given by $R=\frac{V^2\sin \left(2\theta \right)}{g}$ where $g$ is a positive constant, assume $V=2m/s,g=10m/s^2$ and $\theta$ was measured to be $\frac{\pi }{12}$ radians. If there was a possible error in the measurement of $\theta$ of $\frac{1}{10\sqrt{3}}$ radians, estimate the corrosponding error in the computation of the range.

Q. A projectile is launched at t=$0$, with velocity u at an angle with the horizontal