Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • A projectile is thrown at an angle beta with vertical

A projectile is thrown at an angle \beta with vertical .It reaches a maximum height H . The time taken to reach the highest point of its path is 

Answers (1)

best_answer

@Harshit

As we have learned

Time of Flight -

Time for which projectile remains in the air above the horizontal plane.

- wherein

t=\frac{2U\sin \Theta }{g}

SO 

Time taken to reach highest point =\frac{t}{2}=\frac{U\sin \beta }{g}

 

 

 H = \frac{u^2\sin ^2 \beta }{2g} \; \; or \; \; (u\sin \beta )^2= 2gH

u\sin \beta = \sqrt{2gH}........(1)

Time taken to reach highest point = u\sin \beta/ g = \sqrt{2H/g}

Posted by

avinash.dongre

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NEET 2026 Registration LIVE: MBBS, BDS application form closes today; online link, exam date, updates
anam-khan

NEET 2026 registration deadline extended to March 11 for MBBS, BDS; exam on May 3
anam-khan

NEET 2026 Registration LIVE: NTA extends MBBS, BDS application deadline to March 11; how to apply

Latest Question