Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
102 views
in Physics by (82.1k points)
closed by
The trajectory of a projectile is given by `y=x tantheta-(1)/(2)(gx^(2))/(u^(2)cos^(2)theta)`. This equation can be used for calculating various phenomen such as finding the minimum velocity required to make a stone reach a certain point maximum range for a given projection velocity and the angle of projection required for maximum range. The range of a particle thrown from a tower is define as the distance the root of the tower and the point of landing.
From a certain tower of unknown height it is found that the maximum range at a certain projection velocity is obtained for a projection angle of `30^(@)` and this range is `10sqrt(3)m`. The projection velocity must be
A. `10m//s`
B. `10sqrt(3)m//s`
C. `(10)/(sqrt(3))m//s`
D. `5m//s`

1 Answer

0 votes
by (84.3k points)
selected by
 
Best answer
`tan theta=(u^(2))/(Rg) rArr u=sqrt(Rg tan theta)=10m//s`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...