____​2.​An object is thrown vertically upward with a speed of 25 m/s. How much time passes before it comes back down at 15 m/s? (Air resistance is negligible.)

To solve this problem, we can use the equation of motion for an object in free fall:

v = u + gt

where:
v = final velocity
u = initial velocity
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time

Given:
u = 25 m/s (initial velocity)
v = -15 m/s (final velocity, negative because it is in the opposite direction)
g = 9.8 m/s^2

We can substitute these values into the equation and solve for t:

-15 m/s = 25 m/s + (9.8 m/s^2)t

Rearranging the equation, we get:

9.8 m/s^2 t = -15 m/s - 25 m/s

Combining the terms on the right side:

9.8 m/s^2 t = -40 m/s

Now, divide both sides of the equation by 9.8 m/s^2 to solve for t:

t = -40 m/s / 9.8 m/s^2

Calculating the value:

t ≈ -4.08 s

Since time cannot be negative, we discard the negative sign and take the positive value.

Therefore, the object takes approximately 4.08 seconds to come back down at 15 m/s.