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.
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
1 answer