A 49-kg lead ball is dropped from the leaning tower of Pisa. The tower is 54 m high.
(a) How far does the ball fall in the first 2.4 seconds of flight?
(b) What is the speed of the ball after it has traveled 4.1 m downward?
(c) What is the speed of the ball 1.7 s after it is released?
Galileo says:
Who cares what the mass is ?
How far does anything fall in 2.4 s if dropped on a planet where g = 9.81 m/s^2 ??
d = (1/2) a t^2
d = (1/2)(9.81) (2.4)^2
(b) d = 4.1
4.1 = (1/2) (9.81)(t^2)
solve for t
v = Vi + a t
v = 0 + 9.81 t
(c) v = 0 + 9.81 t = 9.81 (1.7)
Now surely this is in your text ?
To solve these problems, we can use the equations of motion for free fall. The key equation is:
s = ut + (1/2)gt^2
where:
s is the distance fallen
u is the initial velocity (which is 0 for a dropped object)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time elapsed
(a) To find the distance the ball falls in the first 2.4 seconds of flight, we can substitute the values into the equation:
s = 0 * 2.4 + (1/2) * 9.8 * (2.4)^2
s = 0 + (1/2) * 9.8 * 5.76
s = 0 + 4.9 * 5.76
s ≈ 28.224 meters
So, the ball falls approximately 28.224 meters in the first 2.4 seconds.
(b) To find the speed of the ball after it has traveled 4.1 meters downward, we can use the equation:
v^2 = u^2 + 2gs
where:
v is the final velocity (the speed of the ball)
s is the distance fallen
g is the acceleration due to gravity (approximately 9.8 m/s^2)
u is the initial velocity (which is 0 for a dropped object)
We need to solve for v, so we rearrange the equation:
v^2 = 0^2 + 2 * 9.8 * 4.1
v^2 = 0 + 2 * 9.8 * 4.1
v^2 = 0 + 80.36
v ≈ √80.36
v ≈ 8.96 m/s
So, the speed of the ball after it has traveled 4.1 meters downward is approximately 8.96 m/s.
(c) To find the speed of the ball 1.7 seconds after it is released, we can again use the equation:
v = u + gt
where:
v is the final velocity (the speed of the ball)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time elapsed
u is the initial velocity (which is 0 for a dropped object)
Substituting the values:
v = 0 + 9.8 * 1.7
v = 0 + 16.66
v ≈ 16.66 m/s
So, the speed of the ball 1.7 seconds after it is released is approximately 16.66 m/s.