Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A projectile fired with initial velocity   at an angle <img alt="\theta" src="https://entrancecorner.oncodecogs

A projectile fired with initial velocity  v at an angle \theta. If initial velocity be tripled at the same angle of projection, then time of ascent will become 

Option: 1

Three times the original time of ascent


Option: 2

Two times the original time of ascent


Option: 3

Four times the original time of ascent


Option: 4

Five times the original time of ascent


Answers (1)

best_answer

Assuming no air resistance, the time of ascent of a projectile launched at an angle  \theta with initial velocity v is given by:

t = (v*sin\theta) / g

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

If the initial velocity is tripled while keeping the launch angle \theta constant, the new initial velocity is 3v Therefore, the new time of ascent,t', is given by:

t' = (3v*sin\theta) / g

To find the ratio of t' to t, we can divide t' by t:

t'/t = [(3vsin\theta) / g] / [(vsin\theta) / g]

t'/t = 3

Therefore, if the initial velocity is tripled while keeping the launch angle constant, the time of ascent of the projectile will become three times the original time of ascent.

Posted by

mansi

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