A more realistic scenario is having the direction of gravity towards a center, which is definitely much harder to derive such an equation, and also you will have to redefine the distance traveled as Δθr, assuming that Earth is a perfect sphere with radius(r). However, this only works for the scenario that the direction of gravity is always one direction that is vertically downwards. Hence the equation can be simplified to s = v^2sin(2θ)/g. Lets remind us about the trigonometry identity sin(2θ) = 2cos(θ)sin(θ). Initial horizontal velocity is typically written as ux subscript x is used to represent the horizontal rectilinear motion. Subsititing the equation, getting s = 2v^2sin(θ)cos(θ)/g. From the equation s = vcos(θ)t, and t = 2vsin(θ)/g. Rearranging the equation for finding t, vsin(θ)/g = t, this is the time it takes to reach its maximum height, so we multiply by 2 to get the total time for it to reach the maximum height and return back to the initial height. At maximum height, the vertical velocity(vsin(θ)) is reduced to zero, so the equation should give vsin(θ) - gt = 0. Since these two components of motion are independent of each other, two distinctly separate sets of equations are needed - one for the projectile's horizontal motion and one for its vertical motion. Knowing that the time it takes for the projectile to reach the maximum height from its initial height is the same as the time it takes to fall from the maximum height back to its initial height. The above equations work well for motion in one-dimension, but a projectile is usually moving in two dimensions - both horizontally and vertically. So the issue is to find time(t), the time is affected by the vertical component of velocity and the acceleration due to gravity(g). Projectile motion only occurs when there is one force applied at the beginning on the trajectory, after which the only interference is from gravity. The path that the object follows is called its trajectory. Knowing that the horizontal velocity = vcos(θ), so we can get the horizontal distance(s) = horizontal velocity x time, s = vcos(θ)t.Ģ. Projectile motion is a form of motion where an object moves in a bilaterally symmetrical, parabolic path. (b) The horizontal motion is simple, because a x 0 and v x is thus constant. Hence the optimal angle of projection for the greatest horizontal distance is 45° because sin(90) = 1, and any other angle will result in a value smaller than 1.ġ. Figure 3.35 (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. I tried to drive a formula, ending up having the horizontal distance traveled = v^2sin(2θ)/g. For the question of comparing the horizontal distance traveled of different initial angles of projection.
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