A rubber ball is dropped from a height of 1.5m.
a) How long does it take to hit the ground?
b) What is the velocity of the ball immediately before it makes contact with the ground?
a) To find the time it takes for the ball to hit the ground, we can use the equation for free fall:
h = (1/2)gt^2
Where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. Given that h = 1.5 m, we can solve for t:
1.5 = (1/2)(9.8)t^2
3 = 9.8t^2
t^2 = 3/9.8
t^2 ≈ 0.306
t ≈ √0.306
t ≈ 0.553 seconds
So, it takes approximately 0.553 seconds for the ball to hit the ground.
b) The velocity of the ball immediately before it makes contact with the ground can be found using the equation:
v = gt
Where v is the velocity and g is the acceleration due to gravity. Plugging in the values:
v = (9.8)(0.553)
v ≈ 5.42 m/s
Thus, the velocity of the ball immediately before it makes contact with the ground is approximately 5.42 m/s.
To solve these questions, we can use the laws of motion and the equations of motion. Specifically, we will use the equations of motion for a free-falling object near the surface of the Earth, assuming no air resistance.
a) How long does it take to hit the ground?
The time it takes for a falling object to hit the ground can be found using the following equation:
h = 0.5 * g * t^2
Where:
h is the initial height (1.5m in this case),
g is the acceleration due to gravity (which is approximately 9.8 m/s^2 near the Earth's surface),
t is the time it takes for the ball to hit the ground (what we want to find).
Rearranging the equation, we get:
t = sqrt(2h/g)
Substituting the given values, we have:
t = sqrt(2 * 1.5 / 9.8)
≈ sqrt(0.3061)
≈ 0.553 seconds
Therefore, it takes approximately 0.553 seconds for the ball to hit the ground.
b) What is the velocity of the ball immediately before it makes contact with the ground?
To determine the velocity of the ball just before it hits the ground, we can use the following equation:
v = g * t
Where:
v is the final velocity of the ball (what we want to find),
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 (0.553 seconds, as we found in part a).
Substituting the known values, we have:
v = 9.8 * 0.553
≈ 5.4254 m/s
Therefore, the velocity of the ball immediately before it makes contact with the ground is approximately 5.43 m/s.
To calculate the time it takes for the ball to hit the ground, we can use the equation of motion:
h = 0.5 * g * t^2
Where:
h = height of the ball (1.5m)
g = acceleration due to gravity (9.8m/s^2)
t = time taken
To solve for the time (t), we rearrange the equation:
t = sqrt(2h/g)
Plugging in the values, we have:
t = sqrt(2 * 1.5 / 9.8)
t ≈ 0.55 seconds
So, it takes approximately 0.55 seconds for the ball to hit the ground.
To calculate the velocity of the ball immediately before it makes contact with the ground, we can use the equation:
v = sqrt(2 * g * h)
Where:
v = velocity of the ball
g = acceleration due to gravity (9.8m/s^2)
h = height of the ball (1.5m)
Plugging in the values, we have:
v = sqrt(2 * 9.8 * 1.5)
v ≈ 6.43 m/s
Therefore, the velocity of the ball immediately before it makes contact with the ground is approximately 6.43 m/s.