A container of water is knocked off a 10.00–meter–high ledge with a horizontal velocity of 1.00 meters/second. Calculate the time it takes for the container to reach the ground.
distance = 1/2*g*t^2. You know distance and g (9.8 m/s^2). Solve for time. Note that the horizontal velocity has nothing to do with the problem. The horizontal velocity determines how far horizontally from the ledge it lands and not how long it takes to reach the ground.
To find the time it takes for the container to reach the ground, we can use the equation of motion in the vertical direction.
The vertical motion of the container can be described by the equation:
h = ut + (1/2)gt^2
Where:
h = height of the container (10 meters in this case)
u = initial vertical velocity (0 meters/second in this case as the container starts from rest vertically)
g = acceleration due to gravity (approximately 9.8 meters/second^2)
t = time taken to reach the ground (what we want to find)
Since the container has no initial vertical velocity, the equation reduces to:
h = (1/2)gt^2
Substituting the values:
10 = (1/2) * (9.8) * t^2
Now, we can solve this equation to find the value of t.
Dividing both sides of the equation by (1/2) * 9.8:
t^2 = 10 / (1/2) * 9.8
t^2 = 20 / 9.8
Taking the square root of both sides:
t = sqrt(20 / 9.8)
Calculating the value:
t ≈ 1.43 seconds
Therefore, it takes approximately 1.43 seconds for the container to reach the ground.