Kinematics Question 677
Question: A bomber plane moves horizontally with a speed of 500 m/s and a bomb released from it, strikes the ground in 10 sec. Angle at which it strikes the ground will be $ (g=10m/s^{2}) $
Options:
A) $ {{\tan }^{-1}}( \frac{1}{5} ) $
B) $ \tan ( \frac{1}{5} ) $
C) $ {{\tan }^{-1}}(1) $
D) $ {{\tan }^{-1}}(5) $
Show Answer
Answer:
Correct Answer: A
Solution:
[a]Horizontal component of velocity vx= 500 m/s and vertical components of velocity while striking the ground. $ v _{y}=0+10\times 10=100m/s $
Angle with which it strikes the ground. $ \theta ={{\tan }^{-1}}( \frac{v _{y}}{v _{x}} )={{\tan }^{-1}}( \frac{100}{500} )={{\tan }^{-1}}( \frac{1}{5} ) $
