Question

2 une balle est lancée horizontalement du bord du toit dun immeuble situé à 25,0 m du sol. la vitesse initiale horizontale de la balle est 12,0 m/s. a) calculez le temps mis par la balle pour atteindre le sol.

Explanation:

Step1: Identify the vertical motion formula

The vertical motion of the ball is a free - fall motion. The formula for vertical displacement \(y = v_{0y}t+\frac{1}{2}gt^{2}\). Since the ball is launched horizontally, \(v_{0y} = 0\). The vertical displacement \(y=- 25.0\ m\) (taking downwards as negative) and \(g=-9.8\ m/s^{2}\) (acceleration due to gravity, direction is downwards). The formula simplifies to \(y=\frac{1}{2}gt^{2}\).

Step2: Solve for time \(t\)

From \(y=\frac{1}{2}gt^{2}\), we can express \(t\) as \(t=\sqrt{\frac{2y}{g}}\). Substitute \(y = - 25.0\ m\) and \(g=-9.8\ m/s^{2}\) into the formula:

$$t=\sqrt{\frac{2\times(- 25.0)}{-9.8}}=\sqrt{\frac{50}{9.8}}$$

$$t\approx2.26\ s$$

Answer:

The time taken for the ball to reach the ground is approximately \(2.26\ s\)