The height `y` and the distance `x` along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by `y = (8t - 5t^2) m` and `x = 6tm`, where `t` is in seconds. The velocity with which the projectile is projected at `t = 0` is.A. `6 m s^-1`B. `8 m s^-1`C. `10 m s^-1`D. `14 m s^-1`

The height `y` and the distance `x` along the horizontal plane of a pr

1.	The height `y` and the distance `x` along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by `y = (8t - 5t^2) m` and `x = 6tm`, where `t` is in seconds. The velocity with which the projectile is projected at `t = 0` is.A. `6 m s^-1`B. `8 m s^-1`C. `10 m s^-1`D. `14 m s^-1`
Answer» Correct Answer - C ( c) `v_x = (dx)/(dt) = 6 and v_y = (dy)/(dt) = 8 - 10 t` Putting `t = 0` (Since we have to find initial velocity) `v_y = 8 - 10 xx 0 = 8` `v = sqrt(v_x^2 + v_y^2)) = sqrt(6^2 + 8^2) =10 ms^-1`.

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