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/s2)

Options:

A) tan1(15)

B) tan(15)

C) tan1(1)

D) tan1(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. vy=0+10×10=100m/s

Angle with which it strikes the ground. θ=tan1(vyvx)=tan1(100500)=tan1(15)



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