A ball rolls off a table moving 3.6m/s in the horizontal direction. The table is 1.2 meters high. Find the amount of time needed for the ball to hit the ground.
Kinematic equation:
d = vi*t + (1/2)a*t^2
Since its moving at 3.6m/s in the horizontal direction, disregard it. Hence initial velocity is 0
1.2m = 0 + (1/2)9.81m/s^2 * t^2
(1.2m / (1/2)9.81 m/s^2) = t^2
(1.2m / (1/2)9.81 m/s^2)^1/2 = t
t = 0.49 s
To find the amount of time needed for the ball to hit the ground, we can use the equation of motion for objects in free fall. The equation is:
h = (1/2) * g * t^2
where h is the height (in this case, 1.2 meters), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Rearranging the equation gives:
t = √(2h/g)
Plugging in the given values:
t = √(2 * 1.2 / 9.8)
Now, let's calculate the result:
t = √(0.2449)
t ≈ 0.494 seconds
Therefore, it will take approximately 0.494 seconds for the ball to hit the ground.
To find the amount of time needed for the ball to hit the ground, we can use the kinematic equation:
h = vi*t + (1/2) * g * t^2
Where:
- h is the height of the table (1.2 meters)
- vi is the initial vertical velocity (0 m/s since the ball is rolling horizontally off the table)
- t is the time we want to find
- g is the acceleration due to gravity (9.8 m/s^2)
Since the initial vertical velocity is 0 m/s, the equation simplifies to:
h = (1/2) * g * t^2
Rearranging the equation to solve for t:
t^2 = (2 * h) / g
t = sqrt((2 * h) / g)
Now we can substitute the values:
t = sqrt((2 * 1.2) / 9.8)
t = sqrt(0.2449)
t ≈ 0.49 seconds
Therefore, it would take approximately 0.49 seconds for the ball to hit the ground.