Question

A stone is projected at an angle $45^{\circ }$ with horizontal. A bee follows the trajectory of the stone at a constant speed equal to the initial speed of the stone. Neglect air resistance in motion of stone. The magnitude of acceleration (in $m{s}^{-2}$ ) of the bee at the topmost point of its trajectory is $9.8\alpha .$ Write the value of $\alpha .$
$(g=9.8m{s}^{-2}$

Answer 1

Answer: 2

Let initial velocity of the stone be $u$
For stone, at the highest point Radius of curvature, $R=\frac{{\left(u/\sqrt{2}\right)}^2}{g}=\frac{u^2}{2g}$
The bee follows the stone's trajectory with the constant speed $u$ , $a=\frac{u^2}{R}=\frac{u^2}{u^2/2g}=2g$
$a=2g=19.6m{s}^{-2}=2\times 9.8$