a ball rolls off a table and falls 0.80 m to the floor, landing with a speed of 4.5 m/s.
1. What was the initial speed of the ball?
2. What initial speed must the ball have if it is
to land with a speed of 5.8 m/s?
3. What is the magnitude of the acceleration
of the ball just before it strikes the ground?
1. Well, gravity must have given the ball a real "kick" to reach a speed of 4.5 m/s after falling 0.80 m! So, the initial speed of the ball must have been quite impressive.
2. To land with a speed of 5.8 m/s, the ball needs to channel its inner Usain Bolt and sprint from the edge of the table. I'm sure if it puts in enough effort, it can achieve that initial speed.
3. Just before the ball hits the ground, it experiences quite the "G-force" from acceleration due to gravity. So, the magnitude of its acceleration would be "Earth-shaking!" But in all seriousness, the acceleration would be approximately equal to the acceleration due to gravity, which is 9.8 m/s².
To answer these questions, we can use the equations of motion.
Let's assume the initial height of the ball above the floor is h and the initial speed of the ball is v.
1. To find the initial speed of the ball:
Using the equation v^2 = u^2 + 2as, where v is the final speed (4.5 m/s), u is the initial speed, a is the acceleration (due to gravity, which is approximately 9.8 m/s^2), and s is the distance traveled (0.8 m).
v^2 = u^2 + 2as
(4.5 m/s)^2 = u^2 + 2(-9.8 m/s^2)(0.8 m)
20.25 m^2/s^2 = u^2 - 15.68 m^2/s^2
u^2 = 35.93 m^2/s^2
u = √(35.93 m^2/s^2)
u ≈ 6.00 m/s
Therefore, the initial speed of the ball was approximately 6.00 m/s.
2. To find the initial speed required to land with a speed of 5.8 m/s:
Using the same equation as before:
v^2 = u^2 + 2as
(5.8 m/s)^2 = u^2 + 2(-9.8 m/s^2)(0.8 m)
33.64 m^2/s^2 = u^2 - 15.68 m^2/s^2
u^2 = 49.32 m^2/s^2
u = √(49.32 m^2/s^2)
u ≈ 7.02 m/s
Therefore, the initial speed required for the ball to land with a speed of 5.8 m/s is approximately 7.02 m/s.
3. To find the magnitude of the acceleration just before it strikes the ground:
The acceleration of the ball just before it strikes the ground is equal to the acceleration due to gravity, which is approximately 9.8 m/s^2.
Therefore, the magnitude of the acceleration of the ball just before it strikes the ground is approximately 9.8 m/s^2.
To answer these questions, we can use the principles of kinematics, specifically the equation of motion for freefall:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration due to gravity
s = displacement
For convenience, let's assume the acceleration due to gravity is approximately 9.8 m/s^2.
1. What was the initial speed of the ball?
Given:
Displacement (s) = 0.80 m
Final velocity (v) = 4.5 m/s
Acceleration due to gravity (a) = 9.8 m/s^2
Using the equation of motion: v^2 = u^2 + 2as, we can rearrange the equation to solve for the initial velocity (u):
u^2 = v^2 - 2as
u^2 = (4.5 m/s)^2 - 2(9.8 m/s^2)(0.80 m)
u^2 = 20.25 m^2/s^2 - 15.68 m^2/s^2
u^2 = 4.57 m^2/s^2
u ≈ √4.57 m^2/s^2
u ≈ 2.14 m/s
Therefore, the initial speed of the ball was approximately 2.14 m/s.
2. What initial speed must the ball have if it is to land with a speed of 5.8 m/s?
Given:
Final velocity (v) = 5.8 m/s
Acceleration due to gravity (a) = 9.8 m/s^2
Displacement (s) remains the same, which is 0.80 m.
Using the equation of motion: v^2 = u^2 + 2as, we can rearrange the equation to solve for the initial velocity (u):
u^2 = v^2 - 2as
u^2 = (5.8 m/s)^2 - 2(9.8 m/s^2)(0.80 m)
u^2 = 33.64 m^2/s^2 - 15.68 m^2/s^2
u^2 = 17.96 m^2/s^2
u ≈ √17.96 m^2/s^2
u ≈ 4.24 m/s
Therefore, the ball must have an initial speed of approximately 4.24 m/s to land with a speed of 5.8 m/s.
3. What is the magnitude of the acceleration of the ball just before it strikes the ground?
The magnitude of the acceleration of the ball just before it strikes the ground is equal to the acceleration due to gravity, which is approximately 9.8 m/s^2. This is because the ball is in freefall, and the only force acting on it is gravity. Therefore, the acceleration of the ball remains constant throughout its fall.