A ball is rolling with an initial velocity of 5.55 m/s over a table moves over the edge and falls to the ground. How long is the ball in the air when the table is 1.11m high.
a. How long is the ball in the air before it hits the ground?
b. How far from the edge of the table does the ball land on the ground?
a. using the free-fall equation ... 0 = -1/2 gt^2 + 1.11 ... solve for t
b. distance = velocity * time ... use time from a.
To determine how long the ball is in the air when the table is 1.11m high, we can use the kinematic equations of motion.
a. To find the time it takes for the ball to hit the ground, we can use the equation:
h = ut + (1/2)gt^2
Where:
- h is the vertical displacement (1.11m in this case)
- u is the initial velocity (5.55 m/s)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time it takes for the ball to hit the ground
Rearranging the equation to solve for t:
1.11 = 5.55t + (1/2)(9.8)t^2
This is a quadratic equation. We can solve it using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
Plugging in the values:
a = (1/2)(9.8) = 4.9
b = 5.55
c = -1.11
t = (-5.55 ± sqrt((5.55)^2 - 4(4.9)(-1.11))) / (2(4.9))
By calculating it, we find two solutions:
t ≈ 0.469 seconds (ignoring negative solution)
t ≈ 1.063 seconds
Thus, the ball is in the air for approximately 0.469 seconds before it hits the ground.
b. To find how far from the edge of the table the ball lands on the ground, we can use the horizontal motion equation:
distance = velocity * time
Since there is no horizontal acceleration, the velocity remains constant. The horizontal velocity is the same as the initial velocity of the ball, which is 5.55 m/s.
distance = 5.55 * 0.469
By calculating it, we find:
distance ≈ 2.598 meters
Therefore, the ball lands approximately 2.598 meters from the edge of the table on the ground.