A pool ball leaves a 0.60 meter high table with an initial velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.
10.5 m/s
To solve this problem, we can use the equations of motion.
Step 1: Determine the time taken for the ball to fall to the ground.
We can use the equation:
h = ut + (1/2)gt^2
where h is the height, u is the initial velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
Substituting the values into the equation, we get:
0.6 = 2.4t + (1/2)(9.8)t^2
Rearranging the equation:
0.6 = 2.4t + 4.9t^2
Setting the equation to zero and rearranging:
4.9t^2 + 2.4t - 0.6 = 0
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
where a = 4.9, b = 2.4, and c = -0.6.
Solving the equation:
t = (-2.4 ± √((2.4)^2 - 4(4.9)(-0.6))) / (2(4.9))
t = (-2.4 ± √(5.76 + 11.76)) / 9.8
t = (-2.4 ± √17.52) / 9.8
t ≈ (-2.4 ± 4.18) / 9.8
This gives two possible values for t:
t₁ ≈ 0.21 seconds (ignoring the negative value)
Therefore, the time required for the pool ball to fall to the ground is approximately 0.21 seconds.
Step 2: Calculate the horizontal distance traveled.
The horizontal distance traveled can be calculated using the equation:
d = ut
where d is the distance, u is the initial velocity, and t is the time.
Substituting the values into the equation:
d = 2.4 * 0.21
d ≈ 0.504 meters
Therefore, the horizontal distance between the table's edge and the ball's landing location is approximately 0.504 meters.
To predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location, we can utilize the equations of motion and the principles of projectile motion.
Firstly, we need to determine the time it takes for the ball to hit the ground. We can use the equation for vertical displacement during free fall:
s = ut + 0.5 * at^2
where:
s = vertical displacement (0.60 m)
u = initial vertical velocity (0 m/s as the ball is not projected upwards)
a = acceleration due to gravity (-9.8 m/s^2, negative because it acts downwards)
t = time
Rearranging the equation, we get:
0.60 = 0 * t + 0.5 * (-9.8) * t^2
0.60 = -4.9 * t^2
Simplifying further gives:
t^2 = 0.60 / 4.9
t^2 = 0.1224
Taking the square root of both sides:
t ≈ √0.1224
t ≈ 0.35 seconds (rounded to two decimal places)
Thus, it takes approximately 0.35 seconds for the pool ball to fall to the ground.
Now, to find the horizontal distance between the table's edge and the ball's landing location, we can use the equation for horizontal distance:
d = ut
where:
d = horizontal distance
u = initial horizontal velocity (2.4 m/s)
t = time (0.35 s)
Substituting the known values:
d = 2.4 * 0.35
d ≈ 0.84 meters
Therefore, the horizontal distance between the table's edge and the pool ball's landing location is approximately 0.84 meters.