Two bodies were thrown simultaneously from the origin: one straight up and the other, at an angle, 60° to the vertical. The initial velocity of each…

Question

Two bodies were thrown simultaneously from the origin: one straight up and the other, at an angle, 60° to the vertical. The initial velocity of each body is equal to 10 m s^(-1) Neglecting the air resistance, the distance between the two bodies after t=2 s is ( Use g=10 m s^(-2))

MARKS Content Team · Accepted Answer

Correct answer: (A) 20 m. In the above case first find out the relative speed of the particle, here, v→_A=v_0j^ and v→_B=v_0cosθi^+v_0sinθj^, ⇒v→_AB=v_0j^-v_0cosθi^-v_0sinθj^ and the relative acceleration, a→_AB=-gj^--gj^=0 so relative displacement, s→_rel=v→_ABt+0=v_0j^-v_0cosθi^-v_0sinθj^t⇒s→_rel=v_0t√(2-2sinθ)=10×2√(2-2sin30°)⇒s→_rel=10×2=20 m

Two bodies were thrown simultaneously from the origin: one straight up and the other, at an angle, $60 °$ to the vertical. The initial velocity of each body is equal to $10 m s^{- 1}$ Neglecting the air resistance, the distance between the two bodies after $t = 2 s$ is ( $Use g = 10 m s^{- 2}$ )

Solution

Relevant Questions

Two bodies were thrown simultaneously from the origin: one straight up and the other, at an angle, 60° to the vertical. The initial velocity of each body is equal to 10 m s-1 Neglecting the air resistance, the distance between the two bodies after t=2 s is ( Use g=10 m s-2)

Solution

Relevant Questions

Two bodies were thrown simultaneously from the origin: one straight up and the other, at an angle, $60 °$ to the vertical. The initial velocity of each body is equal to $10 m s^{- 1}$ Neglecting the air resistance, the distance between the two bodies after $t = 2 s$ is ( $Use g = 10 m s^{- 2}$ )