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.)
a) 18 s
b) 27 s
c) 1.0 s
d) 9.8 s
e) 4.1 s
To solve this problem, we can use the kinematic equation for vertical motion:
v = u + at
where:
v = final velocity (15 m/s)
u = initial velocity (25 m/s)
a = acceleration (which is due to gravity and is approximately equal to -9.8 m/s^2, since the object is moving upwards)
t = time
Rearranging the equation to solve for time, we have:
t = (v - u) / a
Substituting the given values into the equation, we get:
t = (15 - 25) / -9.8
t = -10 / -9.8
t ≈ 1.02 s
Since time cannot be negative, we can discard the negative sign and the solution is approximately 1.02 seconds. Therefore, the answer is option c) 1.0 s.