The Zero Gravity Research Facility at the NASA Lewis...

The Zero Gravity Research Facility at the NASA Lewis Research Center includes a $145 \mathrm{~m}$ drop tower. This is an evacuated vertical tower through which, among other possibilities, a $1 \mathrm{~m}$ 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 reaches a catching device at the bottom of the tower? (c) When caught, the sphere experiences an average acceleration of $25 g$ as its speed is reduced to zero. Through what distance does it travel while stopping? is its initial speed as it leaves the ground? (b) What is its speed at the height of $0.544 \mathrm{~m} ?$ (c) How much higher does it go?

The Zero Gravity Research Facility at the NASA Lewis Research Center includes a $145 \mathrm{~m}$ drop tower. This is an evacuated vertical tower through which, among other possibilities, a $1 \mathrm{~m}$ 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 reaches a catching device at the bottom of the tower? (c) When caught, the sphere experiences an average acceleration of $25 g$ as its speed is reduced to zero. Through what distance does it travel while stopping?
is its initial speed as it leaves the ground? (b) What is its speed at the height of $0.544 \mathrm{~m} ?$ (c) How much higher does it go?

Key Concepts

Uniformly Accelerated Motion

This concept refers to motion in which an object's acceleration remains constant, such as the acceleration due to gravity near Earth. It allows us to use kinematic equations to relate displacement, time, initial velocity, and final velocity. These equations are fundamental in problems involving objects in free fall or moving with a constant acceleration, forming the basis for further analysis in both free-fall and projectile motion situations.

Free Fall

Free fall describes the motion of an object under the sole influence of gravity, without any other forces (like air resistance) acting on it. In such cases, the acceleration is constant and equal to gravitational acceleration. This allows straightforward calculation of the time of fall and the velocity reached by the object when falling from a known height using the kinematic relations.

Deceleration and Stopping Distance

When an object experiences a deceleration—especially a deceleration greater than the acceleration of gravity—the distance it travels while coming to a stop can be calculated using kinematic formulas. This concept is important for understanding how energy is dissipated during rapid stops, and it involves determining the distance over which the object's kinetic energy is reduced to zero under a known negative acceleration.

Projectile Motion

Projectile motion involves objects that are launched into the air and move under the influence of gravity. This concept uses the same kinematic principles as free fall but typically considers an initial upward or angled velocity. Understanding projectile motion allows for the determination of various parameters such as initial speed, speed at a certain height, and maximum height reached during the flight.

Transcript

00:01 Hello and welcome to this video solution of numerate here it's given that the zero gravity research facility at the nasa lewis research center includes a 145 meter drop tower right this is an evacuated vertical tower through which among other possibilities a 1 meter diameter sphere containing a experimental package can be dropped right now a how long is a sphere in free fall right now? here, it's clearly given that it's evacuated right that means it's it's a vacuum right so the main point here is there is no air resistance, right? so this entire 1 meter diameter sphere will suffer no air resistance, so it will fall similarly like other bodies through the space right so here if to calculate the time it will take right so here what you can do is you can apply s equal to u t plus half 80 square right? so here s is the displacement which is this height of 145 meters u is 0 because it started from rest half times acceleration will be due to gravity which is 9 .8, and you have got t square so the time taken will be equal to let's see it's 5 .4 seconds right is a time it will take to reach the ground in free fall now what will be its speed just as it is reaching a catching device at the bottom of the tower right b again the same formula v equal to u plus 80 you will be 0 the acceleration is 9 .8 and this time is 5 .4 right? so this is speed will be pretty large that is 53 .3 meters per second right next c is when caught the sphere experiences an average acceleration of 25 g as its speed is reduced to 0 through what distance it is traveling while it is being stopped right? you can apply the formula v square equal to u square plus 2 a s right from here.

02:29 You can easily calculate the stopping distance final velocity v will be 0 initial speed is the final speed which is 53 .3, right? right and plus 2 times the acceleration will be minus of 25 g the same in negative format because it's a retardation right? it's stopping the ball the sphere sorry, so s is the stopping distance...

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