A marble rolls off the edge of a table that is 0.82 m high. The marble is moving at a speed of 0.11 m/s at the moment that it leaves the edge of the table. How far from the table does the marble land?
how long is it in the air?
.82=1/2 g t^2 solve for time t in the air.
distance=.11m/s * t
3.175
To find the distance from the table where the marble lands, we can use projectile motion equations and consider the vertical motion of the marble.
First, let's consider the equation of motion for the vertical direction:
h = h0 + v0*t + (1/2)*g*t^2
Where,
h is the vertical displacement of the marble (which is the distance from the table where it lands),
h0 is the initial height of the marble (0.82 m in this case),
v0 is the initial velocity of the marble (0.11 m/s in this case),
t is the time of flight (unknown),
g is the acceleration due to gravity (which is approximately 9.8 m/s^2).
We are looking for the value of h, so let's solve for t first.
At the highest point of the marble's trajectory, the vertical velocity component becomes zero (v = v0 + g*t = 0). We can solve this equation for t:
t = -v0/g
Substituting the given values:
t = -0.11 m/s / (-9.8 m/s^2) = 0.0112 s (approx.)
Now, let's find the vertical displacement h using the equation of motion:
h = h0 + v0*t + (1/2)*g*t^2
Substituting the values:
h = 0.82 m + (0.11 m/s)(0.0112 s) + (1/2)(9.8 m/s^2)(0.0112 s)^2
h ≈ 0.82 m + 0.0012 m + 0.00006 m
h ≈ 0.82126 m (rounded to 5 decimal places)
Thus, the marble lands approximately 0.82126 meters from the table.