Projectile problems with solutions are also included in this site. Also an interactive html 5 applet may be used to better understand the projectile equations. We assume that air resistance is negligible and the only force acting on the object is the force of gravity with acceleration g = 9.8 m/s 2. Therefore, maximum height reached = 4.Consider a projectile being launched at an initial velocity v 0 in a direction making an angle θ with the horizontal. (negative as initial vertical velocity is up).Also an interactive html 5 applet may be used to better understand the projectile. We assume that air resistance is negligible and the only force acting on the object is the force of gravity with acceleration g 9.8 m/s 2. It lands at the same height that it was launched. Consider a projectile being launched at an initial velocity v 0 in a direction making an angle with the horizontal. We derive the following equation for the range:Ī projectile is launched at 15 m/s at angle of 40° to the horizontal as shown below. (horizontal vector of initial velocity, ).Using the equation: and writing this with horizontal subscripts: Will it clear the fence The basic motion equations can be solved simultaneously to express y in terms of x. A key point here is that the projectile has a constant horizontal velocity The range of a projectile considers the horizontal part of the projectiles motion. We derive the following equation for the time to reach maximum height: We derive the following equation for maximum height:įor a projectile that starts and finishes its trajectory at the same height the total flight time will be 2× the time the projectile takes to reach its maximum height: (vertical vector of initial velocity, ).(vertical velocity is at maximum height).
Using the equation: and writing this with vertical subscripts: A key point here is that at the maximum height the vertical velocity will be. The maximum height reached considers the vertical part of the projectiles motion. Details of the calculation: The astronauts range is R (v 0 2 sin2 0)/g v 0 2 sin90 o /g v 0. To have maximum range for a given initial velocity, her launch angle must be 0 45 o. The range of a projectile over level ground is R (v 0 2 sin2 0)/g. These variables are often the link to solving more difficult problems consisting of several parts. We have motion with constant acceleration in two dimensions, or projectile motion. The following are common values that may need to be derived in many projectile motion problems: *It does not matter which direction you choose to be positive, both will calculate the same answer if direction is consistent throughout the working. A key result of this is that the acceleration due to gravity will always be positive ( ). All problems analysed here will consider down as positive*. Vertically: As projectiles can move in both directions vertically, a direction (up or down) must be noted as positive.Horizontal: As projectiles will only ever move in one direction horizontally, we naturally make this direction positive.This means that direction is a very important consideration for the analysis of projectile motion problems. Projectile motion deals with many variables which are vectors. The following equations are applied to projectile motion problems:Īs projectile motion problems are analysed in their horizontal and vertical vector components, the equations need to be written with subscripts to reflect this analysis – for example:
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