SATHEE: Kinematics Question 509

Kinematics Question 509

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 $(g = 10 m / s^{2})$ [MH CET 2003]

Options:

A) $\tan^{- 1} (\frac{1}{5})$

B) $\tan (\frac{1}{5})$

C) $\tan^{- 1} (1)$

D) $\tan^{- 1} (5)$

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Answer:

Correct Answer: A

Solution:

Horizontal component of velocity vx= 500 m/s and vertical components of velocity while striking the ground. $v_{y} = 0 + 10 \times 10 = 100 m / s$ Angle with which it strikes the ground.

Kinematics Question 509

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 $(g = 10 m / s^{2})$ [MH CET 2003]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Kinematics Question 509

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) [MH CET 2003]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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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 $(g = 10 m / s^{2})$ [MH CET 2003]