We are told that a man jumps from a plane and we are required to determine the time it takes him to reach 28.5 m/s. Since the man is falling this is a free-fall motion. In free-fall motion, the acceleration is always constant and is equivalent to the acceleration of gravity. This acceleration of symbolized with the letter "g" and its value is:
[tex]g=9.8\frac{m}{s^2}[/tex]Now, to determine the time we will use the following equation of motion for an object in free-fall:
[tex]v_f=v_0-gt[/tex]Where:
[tex]\begin{gathered} v_f=\text{ final velocity} \\ v_0=\text{ initial velocity} \\ g=\text{ acceleration of gravity} \\ t=\text{ time} \end{gathered}[/tex]Since the man jumps from the plane the initial velocity is zero. Therefore, the equation simplifies to:
[tex]v_f=-gt[/tex]Now we solve for "t" by dividing both sides by "-g":
[tex]-\frac{v_f}{g}=t[/tex]Since we need to know the time for the object to reach a velocity of 28.5 m/s, this we take as the final velocity. Since the object is falling the velocity has a negative value. Plugging in the values we get:
[tex]-\frac{(-28.5\frac{m}{s})}{9.8\frac{m}{s^2}}=t[/tex]Solving the operations we get:
[tex]2.9s=t[/tex]Therefore, it takes the man 2.9 seconds to reach the velocity of 28.5 m/s.
