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A rock is launched upward at `45^@`. A bee moves along the trajectory of the rock. What is the magnitude of acceleration `("in" m s^-2)` of the bee at the top point of the trajectory ? For the rock, neglect the air resistance. |
Answer» Correct Answer - `20 ms^-2` Velocity of rock at top point `= u cos theta` So, acceleration of rock =`((u cos theta)^2)/( r) = g` `rArr (u^2)/( r) = (g)/(cos^2 theta) = (g)/(cos^2 45^@) = 20 ms^-2` where `r` is the radius of curvature Now acceleration of bee at top point : `a_b = (u^2)/( r) = 20 ms^-2` Hence, `r` is same for both. |
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