A ball is dropped from rest at the top of a 560 meter tall building. The ball freely falls with constant acceleration for 8 seconds; at this time, the ball reaches a terminal velocity.
a)How far has the ball fallen when it reaches terminal velocity?
b) What is the terminal velocity of the ball?
c)What is the total time that the ball takes to reach the ground?
No way it actually falls with one g while approaching terminal velocity (There is an increasing force up) but I guess we will have to assume that.
Inititial phase
v = - 9.81 t
h = 560 - 4.9 t^2
so when t = 8
v = -9.81*8 = - 78.5 m/s (terminal velocity)
h = 560 - 4.9*64 = 560-314 = 246 meters high (so fell 314 meters so far)
Now part 2
initial height = Hi = 246
constant speed = Vi = -78.5
so 246 meters /78.5 m/s = 3.13 seconds more
8 + 3.13 = 11.1 seconds total
To find the answers to these questions, we need to understand the concepts of free fall and terminal velocity.
a) How far has the ball fallen when it reaches terminal velocity?
When an object is dropped from rest and freely falls under the influence of gravity, it accelerates due to the force of gravity. The distance fallen during this time can be found using the equation for uniformly accelerated motion:
s = ut + (1/2)at^2
where s is the distance fallen, u is the initial velocity (0 m/s for a dropped object), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time. In this case, we need to calculate the distance fallen during the 8-second interval.
s = 0*(8) + (1/2)(-9.8)(8^2)
s = 0 + (-4.9)(64)
s = -313.6 meters
The negative sign indicates that the distance is measured in the opposite direction of the chosen reference point. Therefore, the ball has fallen 313.6 meters downwards when it reaches terminal velocity.
b) What is the terminal velocity of the ball?
Terminal velocity is the constant maximum velocity reached by a falling object when the drag force of the medium it is falling through (such as air) equals the force of gravity. In other words, the net force on the object becomes zero, resulting in no further acceleration.
Since the ball has reached terminal velocity after 8 seconds, we can assume the velocity remains constant from this point onward. Therefore, the terminal velocity is given by the velocity it has reached at this time.
Using the equation of uniformly accelerated motion:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can calculate the terminal velocity:
v = 0 + (-9.8)(8)
v = -78.4 m/s
The negative sign indicates that the velocity is in the opposite direction to the chosen reference point. Therefore, the terminal velocity of the ball is 78.4 m/s downward.
c) What is the total time that the ball takes to reach the ground?
To find the total time it takes for the ball to reach the ground, we first need to determine the time it takes for the ball to reach terminal velocity. Once reaching terminal velocity, it continues to fall at a constant velocity until it hits the ground.
Given that the time taken to reach terminal velocity is 8 seconds, and the total fall time is unknown, we can determine this by rearranging the equation:
s = ut + (1/2)at^2
to solve for time:
t = (√(2s/a))
with s being the distance fallen (560 m in this case) and a being the acceleration due to gravity (-9.8 m/s^2). Plugging in the values:
t = √(2(560)/(-9.8))
t ≈ 10.22 seconds
Therefore, the total time taken for the ball to reach the ground from rest is approximately 10.22 seconds.