Question

A stone is launched upward at ${45}^o$ to horizontal. A bee follows the trajectory of stone at a constant speed equal to initial velocity of projection of stone. The acceleration of the bee at the top most point of trajectory is -

• A. 2g
• B. g
• C. g/4
• D. 3g/2

Answer 1

The correct option is A 2g

at the top most point of the trejactory, velocity of the Stone= vcos$\theta$

where v= initial velocity

$\theta$= angle of projection with horizontal

so, velocity of stone at top most point = vcos45

and since acceleration due to gravity 'g' is acting downward so it will be tangential to velocity at top most point and hence the centripetal acceleration would be = g

Now,

As we know that centripetal acceleration = $\frac{v^2}{R}$ (Where R= radius of curvature)

so by substituting the values we get,

$g=\frac{{\left(v\cos 45\right)}^2}{R}$

$\Rightarrow R=\frac{{\left(v\cos 45\right)}^2}{g}$

Since bee is also following the same path so it's centripetal acceleration would be the same. but it's moving with constant velocity v,

so by again using the formula for centripetal acceleration for be

centripetal acceleration = $\frac{v^2}{R}$

by substituting the value for bee we get,

centripetal acceleration= $\frac{v^{2}g}{{\left(v\cos 45\right)}^2}$ $\Rightarrow$ centripetal acceleration =2g