Home
Class 11
MATHS
A particle just clears a wall of heig...

A particle just clears a wall of height b at distance a and strikes the ground at a distance c from the point of projection. The angle of projection is (1) `tan^(-1)b/(a c)` (2) `"45"^o` (3) `tan^(-1)(b c)/(a(c-a)` (4) `tan^(-1)(b c)/a`

A

4

B

4.91

C

4.19

D

100

Text Solution

Verified by Experts

The correct Answer is:
B
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • MODEL TEST PAPER -3

    ICSE|Exercise Section A |21 Videos
  • MODEL TEST PAPER -3

    ICSE|Exercise Section B |7 Videos
  • MODEL TEST PAPER -3

    ICSE|Exercise SECTION -B|10 Videos
  • MODEL TEST PAPER -2

    ICSE|Exercise Section C |8 Videos
  • MODEL TEST PAPER -4

    ICSE|Exercise SECTION -C|10 Videos

Similar Questions

Explore conceptually related problems

Let a, b and c be positive real numbers. Then prove that tan^(-1) sqrt((a(a + b + c))/(bc)) + tan^(-1) sqrt((b (a + b + c))/(ca)) + tan^(-1) sqrt((c(a + b+ c))/(ab)) =' pi'

If sqrt(3)+i=(a+i b)(c+i d) , then find the value of tan^(-1)(b//a)+tan^(-1)(d//c)

If sqrt(3)+i=(a+i b)(c+i d) , then find the value of tan^(-1)(b//a) + tan^(-1)(d//c)dot

If sqrt(3)+i=(a+i b)(c+i d) , then find the value of tan^(-1)(b//a)tan^(-1)(d//c)dot

If A +B+C= pi and /_C is obtuse then tan A. tan B is

Prove that : tan^(-1)( (a^3 -b^3)/(1+a^3 b^3)) + tan^(-1)( (b^3 - c^3)/(1+b^3 c^3)) + tan^(-1)( (c^3 - a^3)/(1+c^3 a^3)) = 0

The angle between the curves y^2=x and x^2=y at (1,\ 1) is tan^(-1)4/3 (b) tan^(-1)3/4 (c) 90 (d) 45

Prove that tan^(-1)((a-b)/(1+ab))+ tan^(-1)((b-c)/(1+bc))+tan^(-1)((c-a)/(1+ca))=0 , ab>(-1), bc>(-1), ca>(-1)

In a triangle tan A+ tan B + tan C=6 and tan A tan B= 2, then the values of tan A, tan B and tan C are

Which of the following is greatest? tan1 (b) tan4 (c) tan7 (d) tan10