Initial height in horizontal launch

Data

• Initial velocity in the x-axis: v_0x = 230 km/h = 63.8 m/s
• Horizontal distance from the launch: x_f = 250 m

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/s²
• 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:

$$x_f=x_0+v_{0x}·t\Rightarrow 250=63.8·t\Rightarrow t=\frac{250}{63.8}=3.91 s$$ 

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.

$$y_f=H-\frac{1}{2}·g·t^2\Rightarrow H=0+\frac{1}{2}·g·t^2=\frac{1}{2}·9.8·3.{91}^2=\boxed{75.31 m}$$