when the object reaches maximum height, the velocity of the object is 0 m/s.
It reaches maximum height when the gravity of the body has slowed its velocity to 0 m/s. If there is no gravity and there is no external force acting on it then it will never reach a maximum height as there wont be a negativeaccelerationdemonstrated by newtons first law.
Where there is and you have the objects initial velocity then you can use :
v^2 = u^2+2.a.s
v = Velocity when it reaches Max. height so v = 0
u = Initial Velocity (m/s)
a = Retardation/ Negative Acceleration due to gravity, -9.80m/s ^2
And then the unknown (s) is the displacement, or height above ground, and if everything else is in the right format it should be in metres.
when he sits on the thrown
He was thrown out at home.
Fetch means "go and bring back." Catch means "grab ball when thrown."
I think you're referring to the C/C++ concept of "dangling pointers." This is when you allocate some memory to a pointer, then deallocate that memory, but don't change the pointer. This causes any attempted use of the pointer to return an unused memory address. There is no such concept in Java, since the programmer has little to no control over how memory is allocated or freed. The closest thing I can think of is if you're using a class such as a Reader, in which you can close the object (Reader.close()) and then still have a reference to it. But in this case (and other similar cases) attempting to use the Reader further will result in an IOException being thrown.
A heap is unspecified - it is a group of things placed, thrown or lying one on another. It can be a great quantity or a small number. It can refer to a 'heap' of food on a plate to a 'heap' of stone as in a hill. A similar word would be 'pile'
whyh does the sped decreases when an object is thrown vertically up
An object thrown vertically up wards from the ground returned back to the ground in 6s after it was thown up if it reached a height of 12m calculate?
Ignoring air resistance, I get this formula:Maximum height of a vertically-launched object = 1.5 square of initial speed/GI could be wrong. In that case, the unused portion of my fee will be cheerfully refunded.
Acceleration is dependent on the initial velocity of how fast the object is leaving the projectile. The vertical acceleration is greater when the object is falling than when the object reaches the peak in height. However, if the object is thrown horizontally and there is no parabola in its shape then there is not as great of an acceleration.
90
Horizontally
velocity is found by dividing the distance with time. In a second the height traveled is found by multiplying the velocity by the time taken and then dividing the answer by two.
A ball is thrown vertically upward with an initial speed of 20m/s. Two second later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24m/s. At what height above the release point will the ball and stone pass each other?
if the bal is thrown by making 45 degree angles. with the ground..it will travel maximum distance...
If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t^2. If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t^2.
They might seem like they have the same flight patterns, but they don't. A 12-6 curveball thrown by an over-hander will go from 12 to 6, but if the same is thrown by a 3/4 or side-armer, the ball will react a little more like a slider.
apply the formula: v2 - u2 = 2as. Here v= 0; u = 19.6m/s; a = -g ,find s and that's max. heigth