The trajectory of a projectile launched from ground is given by the equation y = -0. The projectile hits the incline plane at point M.Ī) Find the time it takes for the projectile to hit the incline plane.Ī projectile is to be launched at an angle of 30° so that it falls beyond the pond of length 20 meters as shown in the figure.Ī) What is the range of values of the initial velocity so that the projectile falls between points M and N?Ī ball is kicked at an angle of 35° with the ground.Ī) What should be the initial velocity of the ball so that it hits a target that is 30 meters away at a height of 1.8 meters?ī) What is the time for the ball to reach the target?Ī ball kicked from ground level at an initial velocity of 60 m/s and an angle θ with ground reaches a horizontal distance of 200 meters.Ī ball of 600 grams is kicked at an angle of 35° with the ground with an initial velocity V 0.Ī) What is the initial velocity V 0 of the ball if its kinetic energy is 22 Joules when its height is maximum?ī) What is the maximum height reached by the ballĪ projectile starting from ground hits a target on the ground located at a distance of 1000 meters after 40 seconds.ī) At what initial velocity was the projectile launched? ![]() ![]() Problems with Detailed SolutionsĪn object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal.Ī) What is the maximum height reached by the object?ī) What is the total flight time (between launch and touching the ground) of the object?Ĭ) What is the horizontal range (maximum x above ground) of the object?ĭ) What is the magnitude of the velocity of the object just before it hits the ground?Ī projectile is launched from point O at an angle of 22° with an initial velocity of 15 m/s up an incline plane that makes an angle of 10° with the horizontal. An interactive html 5 applet may be used to better understand the projectile equations. These problems may be better understood when 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).Projectile problems are presented along with detailed solutions. 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). 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. 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). At time t, the velocity has two components. The vector initial velocity has two components: V0x and V0y given by: V0x V0 cos () V0y V0 sin () The vector acceleration A has two components A x and A y given by: (acceleration along the y axis only) Ax 0 and Ay - g - 9.8 m/s2. 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 Equations used in the Calculator and Solver. 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. Now the body has two simultaneously independent motions. Point \(O\) is at height \(h\) above the ground. ![]() horizontal projectile motion step by step Projectile Motion. Suppose a body is thrown horizontally from a point \(O\) with velocity \(u\). vertical velocity of projectile using calculator, quickly, based on projectile motion formula. For the question of comparing the horizontal distance traveled of different initial angles of projection. For the derivation of various formulas for horizontal projectile motion, consider the figure given below, The horizontal projection of a projectile.
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