a marble is launched off of a 1.3m high horizontal table with a velocity of 7.2m/sec how far away from the table will the marble land? how long does it takes to hit the floor?
To find out how far away from the table the marble will land and how long it will take to hit the floor, we can use the equations of motion under constant acceleration.
1. Calculate the time it takes for the marble to hit the floor:
The initial height (h₀) of the marble is 1.3m, and the acceleration due to gravity (g) is approximately 9.8m/s² (assuming standard Earth gravity).
We can use the equation of motion: h = h₀ + v₀t - (1/2)gt², where h is the final height, h₀ is the initial height, v₀ is the initial vertical velocity, t is time, and g is the acceleration due to gravity.
Since the marble is launched horizontally from the table, its initial vertical velocity (v₀) is 0 m/s.
Therefore, the equation becomes: h = h₀ - (1/2)gt².
Substituting the values, we get: 0 = 1.3 - (1/2)(9.8)(t²).
Rearranging the equation, we have: (1/2)(9.8)(t²) = 1.3.
Solving for t²:
(1/2)(9.8)(t²) = 1.3
4.9t² = 1.3
t² = 1.3 / 4.9
t² ≈ 0.2653
Taking the square root of both sides, we find:
t ≈ √0.2653
t ≈ 0.515 s (rounded to three decimal places)
Therefore, it takes approximately 0.515 seconds for the marble to hit the floor.
2. Calculate the horizontal distance the marble will travel:
Since the marble is launched horizontally, its horizontal velocity will remain constant throughout the motion. The horizontal velocity is given as 7.2 m/s.
We know that distance (d) is equal to the product of velocity (v) and time (t): d = v * t.
Substituting the values, we get: d = 7.2 * 0.515.
d ≈ 3.708 m (rounded to three decimal places).
Therefore, the marble will land approximately 3.708 meters away from the table.