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.You need to login to perform this action.
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