Mathematically, the height of the object is referred to as hmax. The speed u of an object at (1/2)hmax can be calculated using this equation: v=u*(1-cosq). In this blog post, we will discuss what that means and why it’s important for you to know!
The speed u of an object at the height (0.25)hmax can be calculated using this equation: v=u*(0.75). In this blog post, we will discuss what that means and why it’s important for you to know!
This is a short sentence about how to calculate the speed u of an object at (0.25)hmax mathematically using this equation: v=u*(0.75). In this blog post, we will discuss what that means and why it’s important for you to know!”
What does “the height” mean? hmax or something else?”
In which direction are objects moving when their speeds exceed sound? in paragraph
What is the speed u of an object at (0.25)hmax mathematically?
In what direction are objects moving when their speeds exceed sound? in paragraph
We have discussed a lot so far, but there’s still more to say! We will continue with our discussion tomorrow about how answer these questions and others like them. Tomorrow we’ll talk about: What does “the height” mean? hmax or something else?” In which direction are objects moving when their speeds exceed sound? on paragraph ***
The last sentence should be written as: The next sentences of this blog post will discuss all that has been talked about above plus two new points-speed u at height (0.25)hmax, and how fast the ball would have to be going when it hits the ground in order for this scenario to work.
The next sentences of this blog post will discuss all that has been talked about above plus two new points-speed u at height (0.25)hmax, and how fast the ball would have to be going when it hits the ground in order for this scenario to work.
If we want our object hit a certain velocity v on its way back down, then from equation 12a), we need:
v = sqrt((u*(g))/(h)) wherein g is acceleration due gravity which equals -32 ft/s^(-)(we’ll assume 32ft). This means that if an object wants to hit a velocity of v at the height (0.25)hmax then it would have to be going 32 ft/s on its way up, or if an object wants to hit a velocity of v when it hits the ground again after being thrown up by say (0.75)hmax it would need to also be going 32ft/s-both cases require that u = sqrt(g* h).
For our example case:
the maximum speed we can get for any point along this trajectory is 0.625 ft/sec so 3600 ft/(hr)(hours)/seconds = 360 miles per hour but since there are 60 seconds in one minute and 60 minutes in one hour, this means 180 mph.
For a more thorough explanation of this, see the Wikipedia article on “Maximum Height: Maximum Speed” which is very helpful and links out to other sources.
In our example case where h max = (0.25)hmax then u max = sqrt((g*(0.25))/(0.625)), or 3600 ft/hr * 60 min/hour)/60 secs in one minute=360 mph as mentioned before but since there are 360 degrees in a circle, it means that at least 180 miles per hour will be reached when an object goes up vertically from 0 feet height all the way to its maximum height of (0.75)hmax-see Figure A below for more information about how this works.
Figure A- object 0 ft high and object (0.75)hmax high, the speed of an object is at least 180 mph when it reaches its maximum height in this figure if g=32ft/sqrt(hr). An even more thorough explanation can be found here: Maximum Height: Maximum Speed
As mentioned before, there are 360 degrees in a circle so with 32 feet per second squared for gravity or g then no matter how high up on earth’s surface an object goes to reach its max height that has been given by h max , only about 180 miles per hour will be reached because we know that 120 miles per hours is just under half of our ceiling velocity limitations which means that objects cannot go any faster than that.
For example, if a rocket were to be launched from the surface of earth and it was able to reach an altitude of (0.75)hmax or 25 miles into space, then when returning back down at its max speed, only 180 mph can be reached because our ceiling velocity is 120 mph which means objects cannot exceed this limit by any more than half.
However with winds in the atmosphere affecting gravity’s pull on objects traveling through air as they ascend upwards for height h max , wind resistance will eventually make up for gravitational force enough so that there should be no difference in either an object’s maximum vertical distance traveled over time or its eventual descent downwards upon reaching the peak of potential energy won’t change whether.
Conclusion paragraph:
Understanding this equation can help you figure out how fast your object is traveling at any given point in time. Knowing the speed of an oncoming object could mean the difference between life and death for a cyclist or runner, so it’s important to be aware of what’s happening around you! To find v (speed) use u*(1-cosq). Remember that hmax is height and q is half angle through which we are looking down from above. If you’re wondering where to start with this new knowledge, why not try calculating speeds using different values? You might just get some insight into your own safety habits too!
