question_answer

At the uppermost point of a projectile its velocity and acceleration are at an angle of:

A)  180°                     

B)        90°                        

C)  60°                                        

D)  45°

Correct Answer: B

Solution :

                Key Idea: At the uppermost point of a projectile, the vertical component of velocity becomes zero. When a body is projected at an angle \[\theta \] then it follows a projectile motion. At the highest point P of its path, the vertical component of velocity becomes zero, while horizontal component prevails, also acceleration due to gravity g always acts vertically downwards hence angle between velocity and acceleration is \[=\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}{{k}_{2}}}{({{k}_{1}}+{{k}_{2}})m}}\] Alternative: For a particle projected with velocity u at angle \[=\frac{Force}{Length}=\frac{[ML{{T}^{-2}}]}{[L]}=[M{{T}^{-2}}]\] from the horizontal the height attained is given by \[[ML{{T}^{-2}}]\] From this equation it is clear that for \[=[M{{L}^{-1}}{{T}^{-2}}]\]. \[[M{{L}^{2}}{{T}^{-1}}]\] (maximum), the body will go to maximum height.