The position of a projectile launched from the origin at t = 0 is given by vecr = (40 hati+50 hatj)m at t = 2s. 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 vecr = (40 hati+50 hatj)m at t = 2s. 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

2ux=40 ⇒ 4x=20 50=24y-(1/2)×10×22⇒ 4y = 35 tan θ=(uy/ux)=(35/20)=(7/4) θ= tan-1((7/4))

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

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

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

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

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

$2 u_{x} = 40 \Rightarrow 4 x = 20$
$50 = 24 y - \frac{1}{2} \times 10 \times 2^{2} \Rightarrow 4 y = 35$
$tan θ = \frac{u _{y}}{u _{x}} = \frac{35}{20} = \frac{7}{4}$
$θ = tan^{- 1} (\frac{7}{4})$

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​