Question

A rock is launched upward at ${45}^o$. A bee moves along the trajectory of the rock at a constant speed equal to the initial speed 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.

• A. 20
• B. 30
• C. 40
• D. 50

Answer 1

Answer: (A)

20

velocity of rock at top point =$u\cos \theta$

So, acceleration of rock $=\frac{(u\cos \theta {)}^2}{r}=g$

$\Rightarrow \frac{u^2}{r}$=$\frac{g}{{\cos }^2\theta }=\frac{g}{{\cos }^2{45}^o}=20m{s}^{-2}$

where $r$ is the radius of curvature

Now acceleration of bee at top point

$a_b=\frac{u^2}{r}=20m{s}^{-2}$

Hence $r$ is same for both.