Respuesta :
The speed of the ball just before it hits the ground is given by:
[tex]v=\sqrt[]{v^2_x+v^2_y}[/tex]the horizontal component of the velocity is constant in the complete trajectory, then:
vx = 5 m/s
the vertical component of the velocity is calculated by using the following formula, for the speed in a free fall motion:
[tex]v^2_y=v^2_{oy}+2gy^{}[/tex]voy is the initial vertical component of the velocity and in this case is
voy = 0m/s.
g: gravitational acceleration constant = 9.8m/s^2
y: vertical distance traveled by the ball = 5 m
Solve the previous equation for vy and replace the values of the other parameters:
[tex]\begin{gathered} v_y=\sqrt[]{2gy} \\ v_y=\sqrt[]{2(9.8\frac{m}{s^2})(5m)} \\ v_y\approx9.90\frac{m}{s} \end{gathered}[/tex]Then, by replacing the values of vx and vy, you obtain for v:
[tex]\begin{gathered} v=\sqrt[]{(5\frac{m}{s})^2+(9.90\frac{m}{s})^2} \\ v\approx11.10\frac{m}{s} \end{gathered}[/tex]Hence, the speed of the ball before it hits the ground is approximately 11.10m/s
