SATHEE: Kinematics Question 243

Kinematics Question 243

Question: A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds [AIEEE 2004]

Options:

A) h/9 meters from the ground

B) 7h/9 meters from the ground

C) 8h/9 meters from the ground

D) 17h/18 meters from the ground

Show Answer

Answer:

Correct Answer: C

Solution:

$∵ h = u t + \frac{1}{2} g t^{2}$
$\Rightarrow h = \frac{1}{2} g T^{2}$ After $\frac{T}{3}$ seconds, the position of ball, $h^{'} = 0 + \frac{1}{2} g {(\frac{T}{3})}^{2} = \frac{1}{2} \times \frac{g}{9} \times T^{2}$

$h^{'} = \frac{1}{2} \times \frac{g}{9} \times T^{2}$

$= \frac{h}{9} m$ from top \ Position of ball from ground $= h - \frac{h}{9} = \frac{8 h}{9} m .$

Kinematics Question 243

Question: A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds [AIEEE 2004]

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 243

Question: A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds [AIEEE 2004]

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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