What is the horizontal range of a projectile launched from a height of 250 m with an initial velocity of 50 m/s?

To find the horizontal range of a projectile launched from a height, we need to consider both the vertical and horizontal motion separately, as they are independent of each other.

First, we determine the time it takes for the projectile to fall from the height of 250 meters. We can use the kinematic equation that relates displacement, initial velocity, acceleration due to gravity, and time:

[

y = v_{0y} t + \frac{1}{2} a t^2

]

In this case, the initial vertical velocity ( v_{0y} ) is 0 because the projectile is launched horizontally. The acceleration ( a ) is due to gravity, which is approximately ( -9.81 , \text{m/s}^2 ) (the negative sign indicates that gravity acts downward). The equation thus simplifies to:

[

250 , \text{m} = 0 \cdot t + \frac{1}{2} (-9.81) t^2

]

This can be rearranged into the standard quadratic form:

[

0 = \frac{1}{2} (-9.81) t^2 + 250

]

Solving for (