A ball with a horizontal speed of 1.25 {\rm m/s} rolls off a bench 1.00 {\rm m} above the floor.
How long will it take the ball to hit the floor?
How far from a point on the floor directly below the edge of the bench will the ball land?
Exactly as long as it would take anything else to fall of the bench to the floor.
v = sqrt (2 g h) = sqrt (2*9.8*1) = 4.43 m/s when it hits the floor. Average vertical speed is half that
2.21 m/s average falling speed
1 = 2.21 * time
time = .452 seconds to fall
distance = speed * time
= 1.25 * .452 = .565 meters
To find the time it takes for the ball to hit the floor, we can use the kinematic equation:
h = ut + (1/2)at^2
where:
h = height of the bench (1.00 m)
u = initial vertical velocity (0 m/s, as the ball is rolling off horizontally)
a = acceleration due to gravity (-9.8 m/s^2)
t = time
Since the initial vertical velocity is zero, the equation simplifies to:
h = (1/2)at^2
Plugging in the values, we get:
1.00 = (1/2)(-9.8)t^2
Simplifying, we have:
2 = -4.9t^2
Dividing both sides by -4.9, we get:
t^2 = -0.408
Since time cannot be negative, we discard the negative value. Taking the square root of the remaining value, we have:
t = 0.639 s (rounded to three decimal places)
Therefore, it will take approximately 0.639 seconds for the ball to hit the floor.
To find the horizontal distance from a point on the floor directly below the edge of the bench, we can use the equation:
d = vt
where:
d = horizontal distance
v = horizontal velocity (1.25 m/s, given)
t = time (0.639 s, calculated in the previous step)
Plugging in the values, we have:
d = (1.25 m/s)(0.639 s)
Calculating the value, we get:
d ≈ 0.799 m (rounded to three decimal places)
Therefore, the ball will land approximately 0.799 meters from a point on the floor directly below the edge of the bench.