## Statement

At what height, should you place a cannon capable of launching projectiles at an initial velocity of 230 km/h on the horizontal axis if you would like the projectiles to fall at a distance of 250 meters from where they are shot?

## Solution

**Data**

- Initial velocity in the x-axis:
*v*= 230 km/h = 63.8 m/s_{0x} - Horizontal distance from the launch:
*x*= 250 m_{f}

**Preliminary considerations**

- Horizontal launch is studied as the composition of two motions. A uniform rectilinear motion in the
*x-axis*, and a uniformly accelerated rectilinear motion in the*y-axis*(free fall) - We will place the cannon on the y-axis at a height
*H*which is the value we are looking for - The value of the acceleration of gravity is 9.8
*m/s*^{2} - We will follow the normal sign convention for rectilinear motion

**Resolution**

The projectile moves with a uniform rectilinear motion in the *x-axis*. From the expression for the u.r.m. we can determine the time taken by the projectile to reach the *x-coordinate* of the specified point:

$${x}_{f}={x}_{0}+{v}_{0x}\xb7t\Rightarrow 250=63.8\xb7t\Rightarrow t=\frac{250}{63.8}=3.91s$$

From the expressions for the *y-axis* (free fall) we can calculate the initial height at which the barrel must be placed so that exactly 3.91 seconds later it reaches the ground.

$${y}_{f}=H-\frac{1}{2}\xb7g\xb7{t}^{2}\Rightarrow H=0+\frac{1}{2}\xb7g\xb7{t}^{2}=\frac{1}{2}\xb79.8\xb73.{91}^{2}=\overline{)75.31m}$$