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} ) $



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