Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A ball whose kinetic energy is E ,  is projected at an angle of 45^{\circ} to the horizontal. The kinetic energy of the ball at the highest point of its flight will be

Option 1)

E

 Option 2)

E/\sqrt{2}
Option 3)

E/2

 Option 4)

Zero

Answers (2)

best_answer

As we learnt in

Kinetic energy -

k= \frac{1}{2}mv^{2}

- wherein

m\rightarrow mass

v\rightarrow velocity

kinetic Energy is never negative

 

 

Initial kinetic energy E=\frac{1}{2}mv^{2}

Kinetic energy at highest point =\frac{1}{2}m\left (v cos45^{o} \right )^{2}

 

    =\frac{1}{2}\left (\frac{1}{2}mv^{2} \right )=\frac{E}{2}

Correct answer is 3.    


Option 1)

E

This is an incorrect option.

Option 2)

E/\sqrt{2}

This is an incorrect option.

Option 3)

E/2

This is the correct option.

Option 4)

Zero

This is an incorrect option.

Posted by

perimeter

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: Hardwired to excel? Know BTech electrical engineering cut-offs at NITs
anam-khan

JEE Main 2025 chemistry needs conceptual understanding; 5 tips to score high
anam-khan

JEE Main 2025 exam city intimation slip for session 1 soon; paper pattern, exams from January 22

Latest Question