![]() ![]() 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. Projectile motionis the motion of an object thrown or projected into the air, subject only to acceleration as a result of gravity. 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). Subsititing the equation, getting s = 2v^2sin(θ)cos(θ)/g. From the equation s = vcos(θ)t, and t = 2vsin(θ)/g. The Galileo version describes projectile motion as a combination of horizontal motion with a constant velocity, equal to the projectile initial horizontal velocity component, and vertical motion with constant acceleration and an initial velocity equal to the projectiles initial velocity component. 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. Projectile motion is a form of motion experienced by an object or particle (a projectile) that is projected in a gravitational field, such as from Earths. At maximum height, the vertical velocity(vsin(θ)) is reduced to zero, so the equation should give vsin(θ) - gt = 0. 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. 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). ![]() Knowing that the horizontal velocity = vcos(θ), so we can get the horizontal distance(s) = horizontal velocity x time, s = vcos(θ)t.Ģ. 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. ![]() I tried to drive a formula, ending up having the horizontal distance traveled = v^2sin(2θ)/g. So we should only apply them to the motion of the projectile right after it is thrown and right before it hits the ground. Following are the formula of projectile motion which is also known as trajectory formula: Where, V x is the velocity (along the x-axis) V xo is Initial velocity (along the x-axis) V y is the velocity (along the y-axis) V yo is initial velocity (along the y-axis) g is the acceleration due to gravity. This means that the only force acting on it is the force of gravity. For the question of comparing the horizontal distance traveled of different initial angles of projection. The equations that we are using to solve this problem only apply when the projectile is in free fall. ![]()
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