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
80 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.
In the previous problem, what should be the corresponding projection angle.
A. `tan^(-1)(1//2)`
B. `tan^(-1)(1//3)`
C. `tan^(-1)2`
D. `tan^(-1)3`

1 Answer

0 votes
by (84.3k points)
selected by
 
Best answer
`tan theta=(u^(2))/(Rg) =(100)/(5xx10)=2`
`rArr theta= tan^(-1)2`

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

...