After hitting the tree, how far did the ball travel in the sky when Kaushik saw the ball?
The rest of the post is locked. Join Teachoo Black to see the full post.
CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard
Question 2
Question 3
Question 4 Important
Question 5
Question 6 Important
Question 7
Question 8 Important
Question 9
Question 10
Question 11
Question 12 Important
Question 13
Question 14
Question 15
Question 16
Question 17 Important
Question 18
Question 19 [Assertion Reasoning]
Question 20 [Assertion Reasoning]
Question 21 Important
Question 22 Important
Question 23
Question 24 (Choice 1) Important
Question 24 (Choice 2)
Question 25 (Choice 1)
Question 25 (Choice 2)
Question 26
Question 27 Important
Question 28 (Choice 1) Important
Question 28 (Choice 2) Important
Question 29 (Choice 1) Important
Question 29 (Choice 2) Important
Question 30
Question 31
Question 32 (Choice 1) Important
Question 32 (Choice 2) Important
Question 33 (a) Important
Question 33 (b) Important
Question 34 (Choice 1)
Question 34 (Choice 2) Important
Question 35 Important
Question 36 (i) [Case based]
Question 36 (ii) (Choice 1)
Question 36 (ii) (Choice 2) Important
Question 36 (iii)
Question 37 (i) [Case based]
Question 37 (ii) (Choice 1)
Question 37 (ii) (Choice 2)
Question 37 (iii)
Question 38 (i) [Case based]
Question 38 (ii) (Choice 1)
Question 38 (ii) (Choice 2) You are here
Question 38 (iii) Important
CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard
Last updated at April 16, 2024 by Teachoo
The rest of the post is locked. Join Teachoo Black to see the full post.
We need to find Distance the ball travelled after hitting the tree Given that, After hitting the tree, ball flew towards him i.e we need to find FA Now, FA = GB = CB β CG = 80 β CG β΄ We need to find CG In right angle triangle FCG tan C = (ππππ πππππ ππ‘π π‘π πππππ" " πΆ)/(ππππ ππππππππ‘ π‘π πππππ" " πΆ) tan C = (" " πΉπΊ)/πΆπΊ tan 60Β° = 80/πΆπΊ β3 = 80/πΆπΊ CG = ππ/βπ m Thus, GB = CB β CG = 80 β ππ/βπ = 80(1 β π/βπ) m