A ball is thrown straight up at 20.0 m/s. How fast is the ball moving after 2.00 s?
∆y = vit + 1/2at²
∆y = (20)(2) + 1/2(-9.8)(2)²
∆y = 30.2 m
whoops, thought it said distance
vf = vi + at
vf = (20) + (-9.8)(2)
vf = 0.4 m/s
And that one with the truck and the car bothered you ?
I bet her class uses g = -10 m/s^2 so it came out zero.
To answer this question, we can use the laws of motion and kinematics. The initial upward velocity of the ball is 20.0 m/s, and we need to find its velocity after 2.00 seconds.
First, let's consider that the acceleration due to gravity is always acting on the ball, which is approximately 9.8 m/s² downwards. As the ball moves upward, this gravitational acceleration slows it down until it reaches its highest point, and then the ball falls back down.
To find the velocity of the ball after 2.00 seconds, we'll use the following kinematic equation:
v = u + at
where:
v is the final velocity of the ball
u is the initial velocity of the ball
a is the acceleration
t is the time taken
In this case, the initial velocity, u, is 20.0 m/s, the acceleration, a, is -9.8 m/s² (negative because it's directed downwards), and the time, t, is 2.00 seconds.
Plugging in these values, we have:
v = 20.0 m/s + (-9.8 m/s²) × 2.00 s
v = 20.0 m/s - 19.6 m/s
v = 0.4 m/s
Therefore, the ball is moving downward with a speed of 0.4 m/s after 2.00 seconds.