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# Parabolic Trajectory Calculator

When an object is launched close to the surface of the Earth and the drag force is ignored, the trajectory of the object follows the shape of a parabola. If the initial height of the object is *h _{0}*, the initial velocity

*v*, and the trajectory angle θ, then the parametric equations of the parabola are

_{0}*y*(

*t*) =

*h*+ (sinθ)

_{0}*v*- (

_{0}t*g*/2)

*t*

^{2}

*x*(

*t*) = (cosθ)

*v*,

_{0}twhere

*y*(

*t*) is the y-coordinate of the object at time

*t*,

*x*(

*t*) is the x-coordinate of the object at time

*t*, and

*g*is the gravitational acceleration constant 9.8 m/s

^{2}. See figure below.

### How to Derive These Equations

Since the velocity is a vector, it must be broken down into its x-components and y-components. The velocity in the x-direction is (cosθ)*v*and the velocity in the y-direction is (sinθ)

_{0}*v*. The horizontal trajectory of the object maintains constant velocity since there are no other horizontal forces.

_{0}The vertical trajectory of the object is subject to deceleration due to gravity, hence the -(

*g*/2)

*t*

^{2}term. The reason for dividing

*g*by 2 is so that when you take the second derivative of

*y*(

*t*) you obtain

*g*.

### Critical Points

To calculate when the object reaches its maximum height, set*y'*(

*t*) = 0 and solve for

*t*. If you plug this value into the equation

*y*(

*t*), you can find the maximum height in meters.

To calculate when the object reaches the ground, set

*y*(

*t*) = 0 and solve for

*t*. The larger of the two solutions gives you the time when the object's height is 0. If you plug this value into the equation

*x*(

*t*), you can find how far away the object lands, i.e., its total horizontal displacement.

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