____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 find the time it takes for the object to come back down, we can use the equations of motion.
Let's break down the information given:
Initial velocity (u) = 25 m/s (thrown upwards)
Final velocity (v) = -15 m/s (coming downwards)
Acceleration (a) = -9.8 m/s² (assuming it's on Earth and neglecting air resistance)
We can use the equation:
v = u + at
Rearranging the equation, we get:
t = (v - u) / a
Substituting the values, we have:
t = (-15 - 25) / -9.8
t = -40 / -9.8
t ≈ 4.08 seconds
The object will take approximately 4.08 seconds to come back down at 15 m/s.