The position of a projectile launched from the origin at t=0 is given by hatr=(40 hati+50 hat j) m at t=2 s. If the projectile was launched at an angle θ from the horizontal, then θ is (. take . g=10 ms -2)

Question

The position of a projectile launched from the origin at t=0 is given by hatr=(40 hati+50 hat j) m at t=2 s. If the projectile was launched at an angle θ from the horizontal, then θ is (. take . g=10 ms -2) (A)  tan -1 (2/3) (B)  tan -1 (3/2) (C)  tan -1 (7/4) (D)  tan -1 (4/5)

Tardigrade Admin · Accepted Answer

From question, Horizontal velocity (initial), ux=(40/2)=20 m/s Vertical velocity (initial), 50=uy t+(1/2) g t2 ⇒ uy × 2 (1/2)(-10) × 4 or, 50=2 uy-20 or uy=(70/2)=35 m/s ∴ tan θ=(uy/ux)=(35/20)=(7/4) ⇒ Angle θ= tan -1 (7/4)

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

$tan^{- 1} \frac{2}{3}$

$tan^{- 1} \frac{3}{2}$

$tan^{- 1} \frac{7}{4}$

$tan^{- 1} \frac{4}{5}$

Q. The position of a projectile launched from the origin at t=0 is given by r^=(40i^+50j^​)m at t=2s. If the projectile was launched at an angle θ from the horizontal, then θ is ( take g=10ms−2)

tan−132​

tan−123​

tan−147​

tan−154​

From question, Horizontal velocity (initial), ux​=240​=20m/s Vertical velocity (initial), 50=uy​t+21​gt2 ⇒uy​×221​(−10)×4 or, 50=2uy​−20 or uy​=270​=35m/s ∴tanθ=ux​uy​​=2035​=47​ ⇒ Angle θ=tan−147​

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

$tan^{- 1} \frac{2}{3}$

$tan^{- 1} \frac{3}{2}$

$tan^{- 1} \frac{7}{4}$

$tan^{- 1} \frac{4}{5}$