a 9 kg bowling ball rolls off a table and strikes the ground. if the ball isi n the air for 0.5 seconds, how fast is the ball moving when it hits the ground?
One of the formulas for uniformly accelerated motion (UAM):
h = vo*t - (1/2)gt^2
Ball falls freely, so initial velocity (vo) is zero:
h = 0 - (1/2)(-9.8)(0.5^2)
h = 1.225 meters
Since we're looking for the terminal velocity (vf) we can use the formula,
vf^2 - vo^2 = 2gd
vf^2 - 0 = 2*9.8*1.225
vf^2 = 24.01
vf = 4.9 m/s
To find the speed of the ball when it hits the ground, we need to use the equation of motion:
v = u + at
Where:
v = final velocity (speed of the ball when it hits the ground)
u = initial velocity (initial speed of the ball when it rolls off the table, which we assume to be 0 m/s as it starts from rest)
a = acceleration (due to gravity, which is approximately 9.8 m/s^2)
t = time (0.5 seconds)
Substituting the given values into the equation:
v = 0 + (9.8 m/s^2)(0.5 s)
v = 0 + 4.9 m/s
Therefore, the speed of the 9 kg bowling ball when it hits the ground is approximately 4.9 m/s.
To determine the speed of the bowling ball when it hits the ground, we can use basic kinematic equations. First, we need to find the initial vertical velocity of the ball when it rolls off the table.
Since the ball is in free fall, we can use the equation:
h = vit + (1/2)gt^2
Where:
h is the vertical displacement
vi is the initial vertical velocity
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time (0.5 seconds)
Assuming the ball starts from rest on the table, the initial vertical velocity becomes 0. Therefore, the equation simplifies to:
h = (1/2)gt^2
Plugging in the values, we have:
h = (1/2) * 9.8 * (0.5)^2
h = 1.225 meters
The vertical displacement of the ball is 1.225 meters. Now let's find the final vertical velocity when the ball hits the ground.
Using the equation:
vf = vi + gt
Since the ball starts from rest, the initial vertical velocity (vi) is also 0. Thus, the equation simplifies to:
vf = gt
Plugging in the values:
vf = 9.8 * 0.5
vf = 4.9 m/s
Therefore, the speed of the ball when it hits the ground is 4.9 m/s.