The range of projectile is maximum when the angle of projection is 45 Degrees.
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The [horizontal] range of a projectile is maximised when it shoots at a 45 degree angle. This is true if air resistance is ignored so that the only force acting on the projectile is gravity.
Theoretically, ignoring air friction, 45 degrees. But that also depends on muzzle velocity. If you can fire a projectile at a speed great enough, it would become an orbiting 'satellite' if fired parallel to the ground.
No. An obtuse angle has a measure in the range (90, 180) degrees. The angle in question is greater than the maximum value for an obtuse angle.
vx = 30.0 * cos 60 = 15.0 m/s vy = 30.0 * sin 60 = 25.9 m/s ax = -9.81 m/s/s; ay = 0 m/s/s At the maximum height, vf = 0 m/s 0 = 25.92 - 19.62dy dy = 34.1 m Range I don't know
When a projectile is shot at ground level the range formula is this... d = (v2/g) * sin 2*theta d=range v=velocity g=speed of gravity sin= sine function theta=angle that you make when the projectile is shot In your question you are asking about theta. The maximum distance you will get is when theta is 45 degrees. This is because of gravity. Gravity is pulling down on the object. When something is shot above 45 degrees gravity pull down on the object the same way as it pulls down on the corresponding angle below 45 degrees. When I say corresponding angle look below. Basically the answer to your question is because of gravity. Corresponding angles... 44 and 46 50 and 40 10 and 80 15 and 75 89 and 1 30 and 60 90 and 0 If you notice they all add up to 90 degrees. 90 degrees is the highest angle that can be shot. If it is greater than 90 degrees then the projectile will go behind you.