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What is the equation to calculate the maximum height of an object given an initial velocity and initial launch height?
Wiki User
∙ 12y ago
height=acceletation(t^2) + velocity(t) + initial height
take (T final - T initial) /2 and place it in for time and there you go
Wiki User
∙ 12y ago
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What is the maximum height a sphere shot from cannon can reach?
That depends on basically three things: the angle at which the cannon is fired, the velocity of the projectile, and the acceleration of gravity. The maximum range of the cannon is achieved with a firing angle of 45 degrees to the horizontal. But the maximum height is achieved when the firing angle is 90 degrees to the horizontal, that is, when the cannon is pointing straight up into the air. (Which can be very dangerous to the artillery personnel firing the guns!) So, if we assume that the cannon is on Earth, not some other planet, and fired straight up into the air, can we determine how high the cannonball will go? Yes, if we know the velocity of the cannonball as it leaves the end of the cannon's tube. Once the cannonball leaves the end of the tube, it begins to slow down because of the acceleration of gravity. We can use the energy equations to calculate the maximum height of the cannonball. We know that kinetic energy is defined by the equation Ek = mv2/2. We also know that potential energy (due to altitude) is defined by the equation Ep = mgh. Equating the two, we get mv2/2 = mgh. Rearranging the terms to solve for h, we get: h = v2/2g. (Note that g = 9.8 m/s2 = 32.2 ft/s2.) So, let's say the cannon has a muzzle velocity of 1000 meters per second. The cannonball, therefore, has an initial velocity of 1000 m/s before it starts to slow down. Plugging 1000 into the equation above and solving for h, we calculate the theoretical maximum height as 51,020 meters. In practice, however, the cannonball will not achieve anywhere near that height because of air resistance, which has a tremendous effect at such high speeds.
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What is A football kicked off at an angle 20 from the ground reaches a maximum height of 4.7 m What is the initial velocity of the kick?
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A boy kicks a ball vertically upwards with an initial velocity of 12m/s. Calculate the time taken by the ball to reach the maximum hight and the maximum, height reached by the ball?
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If the launch angle is 15 degrees and the initial velocity is 50.0 meters per second what is the range?
If the initial velocity is 50 meters per second and the launch angle is 15 degrees what is the maximum height? Explain.
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E=mc2 is derived from the equation for kinetic energy Ke = mv2. The mathematics and concepts of special and general relativity shows that the absolute maximum velocity anything can have is the speed of light. The maximum amount of energy anything can possess is simply calculated from its mass and this maximum velocity squared.
A flea jumps straight up to a maximum height of 0.490m What is its initial velocity as it leaves the ground?
At its max height 0.490m its velocity is 0 therefore we can use the equation of motion that v^2 = u^2 + 2as u^2 = 0 - 2 x -9.80(negavtive for direction) x 0.490 therefore u^2 = 7.644 = root 7.644 initial velocity = 2.76 m/s to 2 decimal places hope it helps To calculate the time you can use another equation of motion t = (v-u)/a substitute in : 0-2.76/-9.8 = 0.28seconds but that's only half the journey to its maximum point therefore the time to reach back to the starting point is double this as the acceleration is the same so the answers 0.56 seconds - im sure lol
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If you know an objects mass and the force acting upon it you can calculate its acellaration but can you also calculate its maximum velocity If so whats the formula for this?
V=at. If you figure out the velocity of an object accelerating at 1 g for a year, you will get a velocity that is almost the speed of light (non-relativistically of course).
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When throwing a ball straight up when in the air with an initial velocity of 10 meters per second what will be it maximum height?
'Maximum height' means the exact point at which the velocity changes from upward to downward. At that exact point, the magnitude of the velocity is zero. It doesn't matter what the velocity was when it left your hand. That number determines the maximum height, but the velocity at that height is always zero. --------------------------------------------------------- Thus using the formula: (vf)e2 = (vi)e2+2*a*d vf = final velocity = 0 m/s vi = initial velocity = 10 m/s a = acceleration = gravity = - 9.81 m/s/s d = displacement (distance) = ? e is designating that the next figure is an exponent in the formula So the formula is: (0)e2 = (10)e2 + (2 * -9.81 * d) 0 = 100 + -19.62d adding 19.62d to both sides of the equation 19.62d = 100 dividing by 19.62 d = ~ 5.097 meters
An object is thrown vertically upward when does it reach maximum height?
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.sv = Velocity when it reaches Max. height so v = 0u = Initial Velocity (m/s)a = Retardation/ Negative Acceleration due to gravity, -9.80m/s ^2And 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.
