A 145. drop tower. This is an evacuated vertical tower through which , among other possibilities, a 1m diameter sphere containing an experimental package can be dropped.
a) How long is the sphere in free fall?
b)What is its speed just as it reached a catching device at the bottom of the tower?
c)When caught, the sphere experiences an average deceleration of 25g as its speed is reduced to zero. Through what distance does it travel during the deceleration?
I completly don't understand this question.
which part don't you understand?
For (a), write equation for height (Y) vs. time and solve for the time t when Y=0.
For (b), use V = g t
For c, use the ground-impact velocity V that you get in (b). To decelerate from that velocity within a distance X,
V = sqrt (2 a X)
where a = 25 g
X = V^2/(2 a)
In order to answer these questions, we need to understand the physical principles involved and use appropriate equations. Let's break it down step by step.
a) How long is the sphere in free fall?
To find the time the sphere is in free fall, we can use the equation for the height (Y) of an object in free fall as a function of time (t). The equation is:
Y = 1/2 * g * t^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the height of the tower is not given, but we can assume it is 145 meters. Setting Y to zero (since the sphere reaches the bottom of the tower), we get:
0 = 1/2 * g * t^2
Now we can solve for t:
t^2 = 0 / (1/2 * g)
t^2 = 0
t = 0 seconds
So, the sphere is in free fall for zero seconds. Essentially, the sphere is dropped and immediately reaches the catching device at the bottom of the tower.
b) What is its speed just as it reached a catching device at the bottom of the tower?
To find the speed of the sphere when it reaches the catching device, we can use the equation of motion in the vertical direction:
V = g * t
where V is the velocity (speed) of the sphere and g is the acceleration due to gravity.
Since the time of free fall is zero seconds (as we found in the previous question), the speed of the sphere when it reaches the catching device is also zero. This means it has no speed as it reaches the bottom.
c) When caught, the sphere experiences an average deceleration of 25g as its speed is reduced to zero. Through what distance does it travel during the deceleration?
To find the distance (X) the sphere travels during the deceleration, we can use the equation of motion in the vertical direction:
V^2 = 2 * a * X
where V is the initial velocity of the sphere (which is zero in this case), a is the deceleration and X is the distance traveled during deceleration.
Given that the deceleration is 25g and the initial velocity is zero, we have:
V^2 = 2 * (25g) * X
Since the velocity is zero (V=0) when the sphere is caught, we can simplify the equation to:
0 = 2 * (25g) * X
Now we can solve for X:
X = 0 / (2 * (25g))
X = 0
Therefore, the distance traveled during the deceleration is zero. The sphere comes to a stop immediately as it is caught by the catching device.
I hope this helps! Let me know if you have any further questions.