Feb 27, 2007 · Roughly speaking, if you’re launching something straight up into space, the kinetic energy at the moment of maximum speed (known as burnout speed), just as the engines cut out, is equal to the potential energy at the maximum height of the rocket’s flight. In other words, KEburnout=PEmaximum height.
Put the Model Time on 10ms and press Calculate. The initial result should be close to 180 feet (178.8). Press the 3D Graphs button. Make sure that the Mass of Rocket Empty and Mass of Water option buttons in the Variables (X and Y) frame and Maximum Height in the Results (Z) frame. Click on the Number of Steps combo box and press . Repeat for ...
Then use that estimate to calculate the acceleration in (ft/s2 and m/s2) of an ordinary car. (b) How far (in feet and meters) does the car travel while it accelerates, assuming that the acceleration is uniform? (a) (b) 2.45. Two rockets having the same acceleration start from rest, but rocket A travels for twice as much time as rocket B. (a)
Plugging in what you know — v f is 0 meters/second, v i is 860 meters/second, and the acceleration is g downward (g being 9.8 meters/second 2, the acceleration due to gravity on the surface of the Earth), or -g.You get this: Whoa! The ball will go up 38 kilometers, or nearly 24 miles. Not bad for a birthday present. Theoretically, that 10kg (about 22 lb.) cannonball will come back down and ...
A rocket is traveling straight up at 80 m/s when it runs out of fuel at a height of 7,000 m above the ground. Determine the maximum height reached by the rocket and the time it takes before it falls back down to the ground.Neglect the effect of air resistance.
A rocket engine produces 500 N of thrust. The total mass of the rocket is 40 kg. At one point during its flight, the frictional forces acting on the rocket are 50 N. Calculate: ! (a) the weight of the rocket ! (b) the unbalanced force acting on the rocket ! (c) the acceleration of the rocket. Our Dynamic Universe - Forces, energy and power 3.
Jan 05, 2016 · Use the formula h(t) = —16t2 vt+s to find how long it will take for the ball to hit the ground. Sec. 8. One of the games at a carnival involves trying to ring a bell with a ball by hitting a lever that propels the ball into the air. The height of the ball is modeled by the equation h(t) = — 16t2 +39t.
a) To find the distance the rocket rises in the first 3.2 seconds (from Point 1 to Point 2), relate position and time: equation 1 with a = 27 m/s 2. b) To find the distance the rocket rises between Point 2 and its maximum height (y-velocity = 0), relate position and velocity: equation 3 with a = -9.8 m/s 2.
Then use that estimate to calculate the acceleration in (ft/s2 and m/s2) of an ordinary car. (b) How far (in feet and meters) does the car travel while it accelerates, assuming that the acceleration is uniform? (a) (b) 2.45. Two rockets having the same acceleration start from rest, but rocket A travels for twice as much time as rocket B. (a)
a) Calculate the velocity of the rocket when the the engine stops firing after 5.0s. Vf = (39.2) (5) Vf = 196m/s b) Calculate the maximum height of the rocket. Δd = 0*5 + (1/2) (39.2) (5) 2
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  • My class recently done a water rocket project. Now we have to do a report and i need to calculate the maximum height of our rocket with the use of a inclinometer. At lanch day the angle was 60 and the distance from the person and tyhe lanching pad was 20m. also the height of the person measuring is 5'5.
  • Δ T = M L E V F ( 1 − e − Δ V E V) Where: Δ T: Length of burn in seconds. M L: Total mass of the rocket at the beginning of the burn (often written m 0) E V = Exhaust Velocity in meters/second (often written as v e ). F: Thrust of the rocket in Newtons. Δ V = Delta-V of burn in meters/second.
  • 13. The graph below shows the velocity of a rocket as a function of time. (a) What has happened at point X? [1] (b) When is the acceleration of the rocket a maximum? Suggest an explanation of this. [2] (c) Describe the subsequent motion of the rocket. [2] (d) How in principle would you determine from this graph the maxi-mum height reached by ...

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maximum height . reached by the projectile is shown as . H, and the . launch angle . is marked as . θ. The initial velocity of the projectile is represented as V. o. For all of the following questions, we will assume that the initial height is 0, that is the object is being thrown from the ground level.

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(b) What is the maximum height reached by the rocket? (c) How long after liftoff does the rocket reach its maximum height? (d) How long is the rocket in the air? 54. W A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.00 s for the ball to reach its maximum height.

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Here v_2 = 0 (at max height the rocket is neither rising nor falling), v_1 is the initial vertical component of velocity, and a is -g. This gives you the height it rises above the ramp - you should add the height of the ramp to get the total altitude above the ground. 3. Find the time it takes to reach max altitude, then fall from that altitude.

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A rocket, initially at rest on the ground, accelerates straight upward with a constant acceleration of 34.3 m/s^2. The rocket accelerates for a period of 10.0 s before exhausting its fuel. The rocket continues its ascent until its motion is halted by gravity. The rocket then enters free fall. Find the maximum height, ymax, reached by the rocket.


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2) A toy rocket blasts off from the ground with an initial velocity of 18.0 m/s. Ignoring air resistance, what is the maximum height reached by the rocket before it begins falling? Answer: The first steps in a one-dimensional kinematics problem are to identify what values are known, and then determine which formula will be the most helpful.

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Of course, if we wanted to be lazy, we could have followed the "1/2 Rule" – the maximum height a rocket can achieve, fired straight up, is about equal to half its maximum range, fired for distance.

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The numbers under the L2 heading are the heightof the rocket in feet. 3. Press the STATkey, select CALC, and chooseoption number 5: QuadReg. This will bring you back to the homescreen, with QuadRegshowing. Type "L1, L2" by pressing2ndand 1, then comma, then 2ndand 2.

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Aug 28, 2008 · A model rocket is launched straight upward with an initial speed of 40.0 m/s. It accelerates with a constant upward acceleration of 2.50 m/s2 until its engines stop at an altitude of 160 m. (a) What is the maximum height reached by the rocket? ____m (b) How long after lift-off does the rocket reach its maximum height?

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Sep 05, 2010 · S(2) = 64*2 - 16(2^2) = 128 - 64 = 64. Then the maximum height is S = 64. Now the time needed for the rocket to touch the ground is when the height S = 0. ==> S = 64t - 16t^2 = 0. ==> 16t(4- t ...

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A model rocket is launched straight upward with an initial speed of 70.0 m/s. It accelerates with a constant upward acceleration of {eq}\rm 3.00 m/s^2 {/eq} until its engines stop at an altitude ...

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Excel was used to calculate the performance of the rocket from the motor chart values, the calculated drag coefficient, and the weight of the rocket. S b =38.56 meters, and S c =89.6 meters for a total altitude of 128.2 meters or 423 feet.

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Let's review the 4 basic kinematic equations of motion for constant acceleration: s = ut + ½ at^2 …. (1) v^2 = u^2 + 2as …. (2) v = u + at …. (3) s = (u + v)t/2 …. (4) where s is distance, u is initial velocity, v is final velocity, a is accelerat...

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When the projectile reaches its maximum height its vertical component of velocity will be zero. We can use the linear equations of motion to calculate this height, h. From $v^2 = u^2 + 2 a s$ where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration in this case provided by gravity and $s$ is the distance covered or in our case the height $h$.

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The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot. y=-16x^2+160x+150 Answers: 3

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First, you have to determine how long it will take for the cannonball to reach its maximum height. You know that the vertical velocity of the cannonball at its maximum height is 0 meters/second, so you can use the following equation to find the time the cannonball will take to reach its maximum height: v f = v i + at

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(ii) As the rocket moves upwards, it gains gravitational potential energy. State the maximum gravitational potential energy gained by the rocket. Ignore the effect of air resistance. Maximum gravitational potential energy = _____ J (1) (iii) Calculate the maximum height the rocket will reach. Ignore the effect of air resistance.

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Jul 01, 2013 · A very common application is determine the maximum height of a model rocket launch using the "tangent" function in trigonometry. My first attempt at using this method I used a protractor with a string tied to it and a weight hanging down.

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Oh yes. I know I already did this. However, it was a long time ago with crappy looking graphs. I can do better. The textbooks say that the maximum range for projectile motion (with no air ...

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• The rocket burns 2 tonnes of fuel to accelerate to 1500 m s-1 over a ... Calculate the maximum height reached by the bolt ... if she is to reach the same height

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A rocket is fired vertically from the ground with a vertical acccerleration of 10m/s.The fuel is finished in 1min and it continues to move up.What is the maximum height reached Answer 3 Prakhar

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The height of an object affected by gravity is given by the equation h=(1/2)g*t*t where g=-9.8 meters per second squared or -32 feet per second squared according to the units of the problem. Using right triangle trigonometry and adding the effects of gravity the vertical position of the projectile is represented by the following equation:

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Dec 30, 2019 · Asked on December 30, 2019 by Antara Shahi A 3-kg model rocket is launched straight up with sufficient initial speed to reach a maximum height of 100 m, even though air resistance (a nonconservative force) performs - 900 J of work on the rocket. The height the rocket would have gone without air resistance will be

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minimum of mathematics, yet be able to calculate such things as maximum height H reached above the earth’s surface and the time τ and range L to impact. Our starting point for this discussion is the following schematic showing a typical ICBM trajectory-

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a) Write an equation of the height of . John ’s rocket as time passes, in factored form, , given that . (So,; you find b and c and write them into the equation.) y = -16(x – 0)(x – 3.2) b) At what time on the stopwatch did . Julia ’s rocket reach its maximum height? 5.9 seconds. c) Draw a graph of the height of . Julia ’s rocket, as time passes. Be sure to label the axes carefully.

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Line of sight calculation is based on earth radius and height above ground of transmitter (and receiver). R = 21000000 (21 million feet) approximately. Based on plane trigonometry, d squared = h * (2 * R + h). This is distance to horizon for an observer at height h above surface of radius R.

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What is the maximum height that Alice will reach? So, to find out how high she will jump, Alice needs to find a local maximum of the function \(\text{bounce}(t)\). One way to do this is using calculus. Let's find out how. We'll start out with another example, and come back to Alice later. Example

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A rocket, continuously accelerated by its exhaust, can escape without ever reaching escape velocity, since it continues to add kinetic energy from it engines. It can achieve escape at any speed, given sufficient propellant to provide new acceleration to the rocket to counter gravity's deceleration and thus maintain its speed.

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Actually, the rocket moves faster and faster as the engine is thrusting. At the end of this thrusting portion of the flight (1.7 seconds into flight time from liftoff), the model rocket is traveling at its maximum speed. This maximum speed is 670 feet per second or about 3.5 times as fast as the average speed. After the propellant is gone, the ...

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of July. The rocket height as a function of time is 3x2 + 4x — modeled by the function h(t) —16t2 + 16t + 370, where t is the time in seconds and h is the height in feet. (Use the 'Window' function on your graphing calculator and set the dimensions as shown) a. b. c. How long did it take to reach its maximum height?

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Calculate the maximum height reached by the rocket and how long the rocket is in the air. Projectile Motion: The motion of a particle moving with an initial speed in any direction under the effect ...

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Oct 07, 2019 · To calculate its final position, use the formula s = vt, or use common sense to realize the rocket must be at (5 minutes)(120 meters/minute) = 600 meters north of its starting point. For problems involving constant acceleration, you could solve for s = vt + ½at 2 , or refer to the other section for a shorter method of finding the answer.

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Then use that estimate to calculate the acceleration in (ft/s2 and m/s2) of an ordinary car. (b) How far (in feet and meters) does the car travel while it accelerates, assuming that the acceleration is uniform? (a) (b) 2.45. Two rockets having the same acceleration start from rest, but rocket A travels for twice as much time as rocket B. (a)

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Its height, in feet above the ground, as a function of time, in seconds since it was fired, is given by the equation: h(t) = -16t2 + 112t. (a) At what height was the object fired? (b) Sketch a general curve of this equation below. (c) Algebraically, find the time that the rocket reaches its greatest height and the maximum height.

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Rocket Booster. A small rocket with a booster blasts off and heads straight upward. When at a height of . and velocity of 200.0 m/s, it releases its booster. (a) What is the maximum height the booster attains? (b) What is the velocity of the booster at a height of 6.0 km? Neglect air resistance.


Since the velocity of the ball is $0 \frac{ft}{s}$ when the ball is at its maximum height of $132ft$, you can calculate it starting from there. Since this is more of a physics problem than a math problem, I'm going to do this the more Physics-y way. To make it easier, I'm converting this to meters. $132ft = 40.2336m \approx 40m$.