A ball rolls with a speed of 4.4 m/s toward the edge of a table that is 1 m above the floor. The ball rolls off the table.
a. How long is it in the air? s
b. How far out from the edge of the table does the ball hit the floor? m
The ball after leaving the table
takes part in two motions: horizontal uniform motion
L = v(hor) âąt,
and vertical free fall
H = gt^2/2,
t = sqroot(2H/g) = sqroot(2âą1/9.8) = 0.2 s.
L = v(hor) âąt = 4.4âą0.2=0.88 m.
a ball rolls 5cm in 5 seconds what is the speed of the ball?
To find the answer to these questions, we need to consider that the ball is subject to vertical motion due to gravity while rolling off the table. We can use the equations of motion to solve for the time of flight and horizontal distance traveled by the ball.
a. To determine how long the ball is in the air, we can use the equation of motion for vertical motion:
đŠ = đąđŠđĄ + 0.5đđĄÂČ,
where đŠ is the vertical displacement, đąđŠ is the initial vertical velocity, đĄ is the time, and đ is the acceleration due to gravity.
In this case, đŠ is the height of the table (1 m) and đąđŠ is the vertical component of the initial velocity (which is zero for a rolling ball). We can plug in the values and solve for đĄ:
1 = 0.5 Ă 9.8 Ă đĄÂČ.
Simplifying, we get:
9.8đĄÂČ = 2,
đĄÂČ = 2 / 9.8,
đĄ â â(2 / 9.8).
Evaluating this expression, we find:
đĄ â 0.45 s.
Therefore, the ball is in the air for approximately 0.45 seconds.
b. To determine how far out from the edge of the table the ball hits the floor, we can use the equation of motion for horizontal motion:
đ„ = đąđ„đĄ,
where đ„ is the horizontal distance, đąđ„ is the initial horizontal velocity, and đĄ is the time.
In this case, đąđ„ is the horizontal component of the initial velocity (which is 4.4 m/s). We can plug in the values and solve for đ„:
đ„ = 4.4 Ă 0.45.
Calculating this, we find:
đ„ â 1.98 m.
Therefore, the ball hits the floor approximately 1.98 meters out from the edge of the table.