Horizontal launch of a rolling ball

Data

Height of the ball. H = 2.5 m
Horizontal distance. x = 1 m at t = 0.4 s.
Acceleration of gravity. g = 9.8 m/s²

Preliminary considerations

As described in the statement, the ball behaves as a horizontal launched projectile, so to solve this problem we will need to use the equations for this type of motion.

Resolution

a) Knowing that the ball is on x = 1 m at t = 0.4 s, we will just substitute these values into the equation of the horizontal launch to calculate the velocity of the ball in the x-axis. Remember that in this motion the body moves with uniform rectilinear motion:

$$\begin{gathered} x=x_0+v·t \Rightarrow \\ 1 m = 0 m + v ·0.4 s \Rightarrow \\ v = \frac{1 m}{0.4 s} \Rightarrow \\ \boxed{v = 2.5 m/s} \end{gathered}$$

b) To determine how far the ball will be when it hits the ground we must determine when the impact will occur. To do this, first we determine how long it takes using the equation for position on the y-axis and make y = 0, that is, the ball is on the ground:

$$\begin{gathered} y=H-\frac{1}{2}·9.8 m/s^2·t^2\Rightarrow \\ 0 m = 2.5 m -\frac{1}{2}·9.8 m/s^2·t^2 \Rightarrow \\ {t}^2= \frac{-2.5 m}{-4.9 m/s^2} \Rightarrow \\ \boxed{t = 0.71 s} \end{gathered}$$

Now we only have to substitute the time calculated in the equation of position for the x-axis:

$$\begin{gathered} x=x_0+v·t \Rightarrow \\ x=0m + 2.5 m/s · 0.71 s \Rightarrow \\ \boxed{x=1.77 m} \end{gathered}$$

c) Finally, we can determine the height above the ground the ball is located at t = 0.4 s, using of the equation of position for y for this time:

$$\begin{gathered} y = H - \frac{1}{2}·g·t^2 \Rightarrow \\ y = 2.5 m - \frac{1}{2}·9.8 m/s^2·0.4^2 \Rightarrow \\ \boxed{y = 1.72 m} \end{gathered}$$