initial velocity would be ZERO before launch. To calculate the velocity you would need to hit that target at that distance you would need to know the mass of the rocket and the angle of launch or trajectory simplifying it
If "range" means that the shooter and the target are on the same level: quadrupled (if airesistance can be neglected). It takes twice the time until gravity "eats up" vertical velocity and during that time the projectile moves with double horisontal velocity. But if you shoot horisontally from a cliff at double velocity the flighttime will be the same and the range only doubled.
When the 0.500kg ball collides with the stationary ball, momentum is conserved. Meaning, initial momentum = final momentum. Momentum of an object is = mass(m) x velocity (v). If two objects are in the system, then you have to add up both initial momentums and set them equal to the final momentums... So... m x v(initial, first object) + m x v(initial, second object) = final momentum. (0.500kg)(4.0m/s) + (1.0kg)(0m/s) = final momentum. So the final momentum equals 2.0kgm/s... D. 2.0 kgm/s
It isn't.Gravity is always acting on the bullet, from the time it rests on the breech to the time it hits target, and every point in between.At the beginning of its motion, the bullet has velocity in a particular direction. Gravity accelerates the bullet 'downwards' and alters the velocity at each point in time and space. Eventually, the combination of the previous motion and the present velocity makes it strike a target, at which point forces in the target and bullet result in all motion ceasing, and these forces become balanced.
Basically, a chemical booster rocket propels it into space, once in space the warhead bus separates from the booster rocket, the booster rocket fall toward earth and burns up, the warhead bus maneuvers to aim the warhead(s) at the target(s) and releases the warhead(s). When a warhead arrives at its target at the preset burst height/depth it detonates. There are many other detail steps I have skipped over to keep it simple.
A missile is anything thrown or projected. A rocket uses rocket propulsion - throwing out hot gasses in order to move in the opposite direction. In practice most rockets can be considered to be missiles and most missiles are rockets. in the military a rocket is unguided en a missile is guided, think about GPS or a wire that is in the back of the missile an attached to the launcher an can steered to the target.
A spotting rifle fires a bullet designed to simulate the flight characteristics of the rocket about to be fired. It's basically an aiming device - if the projectile hits the target, then your rocket should hit the target when you fire it.
of course not!
If "range" means that the shooter and the target are on the same level: quadrupled (if airesistance can be neglected). It takes twice the time until gravity "eats up" vertical velocity and during that time the projectile moves with double horisontal velocity. But if you shoot horisontally from a cliff at double velocity the flighttime will be the same and the range only doubled.
The rocket launcher that shoots straight up in the air and comes down vertically on a enemy or target is called a Javelin.
Use a grenade, a rocket launcher & a target locator.
They sell them at staples and possibly target or giant.
Rowling's children
An Initial sales promotion schedule is a schedule that a vendor sets up a schedule to introduce a new product to consumers or clients. It is basically to reinforce loyalty to the brand and keep them relevant. To set up an initial sales promotion schedule goals would be set for what should be achieved and then a target market for the product.
You use it when throwing an object at a target. Over any but a trivially short distance, gravity will pull the object downwards. So you aim higher than the target. To hit the target, the vector sum of the initial velocity and the downward acceleration experienced during the flight must be a vector aimed directly at the target.
They height y of the projectile is given by the function y = vosin(0)t + 1/2gt2, where vo is the initial velocity of the projectile, 0 is the firing angle, t is the time, and g is the acceleration of gravity (-9.81m/s2). The range x of the projectile is given by the function x = vocos(0)t. Rearranging this last equation for time yeilds t = x/(vocos(0)); this will give us the length of time the projectile takes to reach the target. Substituting this into the first equation yeilds: y = vosin(0)[x/(vocos(0))] + 1/2g[x/(vocos(0))]2 this can be simplified further but it is not necessary to do so; plugging it the x and y coordinates, the initial velocity, and the acceleration of gravity, you should be able to solve for 0, which is now the only unknown.
When the 0.500kg ball collides with the stationary ball, momentum is conserved. Meaning, initial momentum = final momentum. Momentum of an object is = mass(m) x velocity (v). If two objects are in the system, then you have to add up both initial momentums and set them equal to the final momentums... So... m x v(initial, first object) + m x v(initial, second object) = final momentum. (0.500kg)(4.0m/s) + (1.0kg)(0m/s) = final momentum. So the final momentum equals 2.0kgm/s... D. 2.0 kgm/s
It isn't.Gravity is always acting on the bullet, from the time it rests on the breech to the time it hits target, and every point in between.At the beginning of its motion, the bullet has velocity in a particular direction. Gravity accelerates the bullet 'downwards' and alters the velocity at each point in time and space. Eventually, the combination of the previous motion and the present velocity makes it strike a target, at which point forces in the target and bullet result in all motion ceasing, and these forces become balanced.