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

`20 ms^-2`

Explanation

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.