A 0.05 kg ball is thrown straight up into the air with an initial speed of 20 m/s. Find the momentum of the ba
at its maximum height its momentum is 0 because the speed is 0 b)KE=PE 0.5(0.05)(20)^2=PE 0.5(0.05)(400)=PE (0.025)(400)=PE 10=PE PE=mgh 10=mgh 10=(0.05)(9.8)h 10=0.49h 10/0.49=h 20.41 meters is the max height so the half height is 10.205 round that to 10.21 halfway up is 10.21 meters now we can use this KE(initial)=KE+PE(final) and the final point is half way up 0.5mv^2=KE+PE(final) KE(initial)=PE+KE(final) 10=PE+KE(final) 10=mgh+0.5mv^2 10=(0.05)(9.8)(10.21)+0.5mv^2 10=5+0.5mv^2 5=0.5mv^2 5=0.5(0.05)v^2 5=(0.025)v^2 200=v^2 14.14=v 14.14 m/s is the velocity halfway up no we can find the momentum p=mv p=(0.05)(14.14) p=0.707 kgm/s
If i throw a ball straight up in the air,what will happen when it reaches its highest point?
When the ball was thrown up, it has kinetic and gravitational energy. When the ball reaches the highest point and 'stops' for a while, all the kinetic energy will change into gravitational energy and has the most gravitational energy due to the height. When the ball falls back due to gravity, the gravitational energy will start to become lesser and it will change back to kinetic energy. ***The gravity will be higher while the speed of the ball will be zero when it reaches the highest point.
It goes up, stops, and comes down. It'll probably land behind the "fast moving object" due to air friction...
When the ball is thrown upwardsFirstly it speed slows down as acceleration due to gravity is decelerating itThe ball reaches a maximum height and again start coming down towards the groundAnd since now the velocity is towards the ground and gravitational acceleration also downwards which then results in imparting greater speed and just before ball hits the ground it's speed is maximum at that point
If a ball is thrown straight up in the air, what is the direction of its acceleration?
thrilling theory. The ball would not come to an abrupt end. in case you measured the fee at which a ball thrown up slows down and reverses, and then plot it on a graph (assuming the verticle axis is velocity and the horizontal axis is time), you will locate it extremely is a suited parabolic curve comparable to a "U" and the backside tip of the "U" touches the horizontal axis (0 velocity). in case you graphed the fee of a fly hitting a convention, even inspite of the undeniable fact that, it would be 3 strains. it would be flat at till a definite element, then slope upward sharply for a cut up 2nd (it extremely is whilst it hits the practice), and the flat back plenty larger up the verticle axis. it would not hit the horizontal axis (0 velocity) in any respect. As for the practice itself, it would decelerate whilst it hits the fly, however the replace in velocity would be so rather small that no person would ever word, no longer in spite of the main state-of-the-artwork methods.
It falls. It falls just like a ball thrown from the top of a skyscraper. The cool part happens when we get to the center of the earth. All of this is assuming no friction and a hole that goes straight through the center of the earth. So, as the ball is increasing its velocity exponentially it's acceleration is actually also increasing because the strength of gravity increases the closer you get to the center. So by the time the ball gets to the center, it is going at its max velocity. The second it passes the halfway point the acceleration vector flips and you now get the ball decelerating in an exactly symmetrical way that if accelerated. With no air resistance the ball is going to oscillate back and forth between the 2 sides of the ball. However, with air resistance, the ball would slow down a little each time and not reach the same height as before. Then, after some time, the ball would stop oscillating right in the center of the earth because gravity is strongest right there and is pushing on the ball from all sides. This is a very fun thought experiment for teaching gravity and acceleration. Enjoy! Here is a fun gif to visualize it. http://en.m.wikipedia.org/wiki/G...
If i throw a ball into the air, what will happen?
The ball will reach a maximum height, then the velocity of gravity (9.8 meters/second) would cause it to travel downward until another force acts upon it (according to Newton's laws of motion). However, this does not account for other forces, such as air resistance. The rate at which the ball falls and the air resistance acting upon it will depend on the properties of the ball (surface area, weight, etc.). If the ball is inside a vacuum, it is said to be in free fall (as the only force acting upon it is gravity) and there is no other force acting upon it. If the ball is allowed to fall long enough, it will reach a point of terminal velocity, where it cannot fall any faster (the force of air resistance will equal the weight of the object)
I’ll go out on a limb and say that I’m the only one who’s going to give you the correct answer.The short answer is that the ball is not a perfectly rigid structure and calculating its exact velocity (including direction) at a given moment is actually impossible.Specifically, anyone who’s giving you an answer that’s focused on the mathematics of the travel of the ball is giving you only an academic answer and thus a wrong answer. Nature does not strictly adhere to mathematical equations, and gravity is a physical phenomenon, not a mathematical one.Gravity occurs because of fields of energy and/or fields of particles. Anyone who disagrees with this would have a tough time providing a legitimate explanation of why gravity appears to propagate at the speed of light. Since gravity is almost certainly particle-based, and a finite number of particles are involved in any gravitational interaction, the acceleration of an object (such as a ball) due to gravity is accomplished in small discrete quanta of particle interactions. Because of this, you might think that the ball stops at zero velocity for a tiny amount of time at the top of its upward trajectory. However, if we were able to discern its exact velocity at all moments, we might find that it never reaches exactly zero velocity, but that with the final particle interaction at the top of its trajectory, it instantaneously switches from a minuscule upward velocity to a minuscule downward velocity.But of course it’s more complex than this. We could try to dissect the event of this final velocity transition to get a more exact answer. It would depend upon the exact interaction of the final particle with the ball. If I had to give an answer to the question of this final transition of velocity, I’d say that it’s not really an instantaneous transition because the ball is not a perfectly rigid object, and that the physical structure of the ball is slightly oscillating in response to each particle interaction. Because the ball is physically oscillating, the answer to your question is that ascertaining the exact velocity of the ball is nondeterministic so we can’t perfectly calculate the ball’s velocity to the point that we can determine when the ball technically starts traveling in the opposite direction.
If you THROW an object downward on earth, what is its acceleration? Why?
If AIR DRAG is neglected, it is EXACTLY 9.8 m/s^2. Gravity is the ONLY force acting on the object in its falling body motion, and it always applies its value of 9.8 Newtons/kilogram (presuming it is on Earth and the standard value is acceptable) of force to the body. With Newton's 2nd law, this corresponds (by an interesting coincidence of mass proportional gravity) to an acceleration of 9.8 m/s^2.