Saved Bookmarks
| 1. |
A ground to ground projectile projected at t=0 is at point A at t=T/3 is at point B at t=5T/6 and reaches the ground at t=T. The difference in heights between point A and B is |
|
Answer» Let heights difference between A and B = H Given :- t = T/3 t = 5T/6 As we know that, Time of Flight, T = 2u Sinθ/g u Sinθ = gT/2 ------(1) For A, h = usinθ t - 1/2 gt² h = usinθ × T/3 - gT²/18 h = gT²/6 - gT²/18 h = 2gT²/18 For B, h' = usinθ t - 1/gt² h' = gT/2 × 5T/6 - 1/2 g 25T²/36 h' = 5gT²/12 - 25gT²/72 h' = 5gT²/72 Again, H = h - h' = 2gT²/18 - 5gT²/72 H = (8gT² - 5gT²)/72 H = 3gT²/72 H = gT²/24 |
|