Answer:
57 meters
Explanation:
Let's see what variables we have in this problem:
- [tex]v_i \ \ \ \ \ \ \ t \\ v_f \ \ \ \ \ \ \triangle x \\ a[/tex]
Let's set the upwards direction to be positive and the downwards direction to be negative.
We are given the initial velocity, 4 m/s. Since the ball is thrown directly downward, we can say that the initial velocity = -4 m/s.
We are also given the acceleration due to gravity, and since the acceleration is pointing downwards, we can say that a = -10 m/s².
The time is also given to us; the question wants to know the vertical displacement at time = 3 seconds, so we can plug in 3 seconds for t.
Since we are solving for vertical displacement, we can use [tex]\triangle x[/tex] as one of our variables.
Now we have all of the variables except for the final velocity, [tex]v_f[/tex].
Since we are dealing with constant acceleration, we can use this constant acceleration equation:
- [tex]x_f=x_i+v_it+\frac{1}{2}at^2[/tex]
Subtract [tex]x_i[/tex] from both sides to get [tex]\triangle x[/tex].
- [tex]\triangle x =v_it+\frac{1}{2}at^2[/tex]
Substitute in the known values:
- [tex]\triangle x = (-4\frac{m}{s})(3s) + \frac{1}{2}(-10\frac{m}{s^2})(3s)^2[/tex]
Get rid of the units to make the equation more readable.
- [tex]\triangle x = (-4)(3) + \frac{1}{2} (-10)(3)^2[/tex]
Simplify the equation and solve for delta x.
- [tex]\triangle x=(-12) -5(9)[/tex]
- [tex]\triangle x =-12-45[/tex]
- [tex]\triangle x = -57[/tex]
The vertical displacement is -57 meters, so we can say that the distance the ball falls in 3 seconds is most nearly 57 meters.
