Question

Please ignore what I began writing and explain in laymen terms for I am in 9th grade.

Answer 1

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:

`g=9.8(m)/(s^2)`

Now, to determine the time we will use the following equation of motion for an object in free-fall:

`v_f=v_0-gt`

Where:

`\begin{gathered} v_f=\text{ final velocity} \\ v_0=\text{ initial velocity} \\ g=\text{ acceleration of gravity} \\ t=\text{ time} \end{gathered}`

Since the man jumps from the plane the initial velocity is zero. Therefore, the equation simplifies to:

`v_f=-gt`

Now we solve for "t" by dividing both sides by "-g":

`-(v_f)/(g)=t`

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:

`-((-28.5(m)/(s)))/(9.8(m)/(s^2))=t`

Solving the operations we get:

`2.9s=t`

Therefore, it takes the man 2.9 seconds to reach the velocity of 28.5 m/s.