Saved Bookmarks
| 1. |
The height `y` and distance `x` along the horizontal plane of a projectile on a certain planet are given by `x = 6t m` and `y = (8t^(2) - 5t^(2))m`. The velocity with which the projectile is projected isA. `8 ms^(-1)`B. `9 ms^(-1)`C. `10 ms^(-1)`D. `(10//3) ms^(-1)` |
|
Answer» Correct Answer - C `v_(y) = (dy)/(dt) = 8 - 10t, v_(x) = (dx)/(dt) = 6` At `t = 0, v_(y) = 8 ms^(-1)` and `v_(x) = 6 ms^(-1)` `:. V = sqrt(v_(x)^(2) +v_(y)^(2)) = 10 ms^(-1)` |
|