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Why does a catapult shoot best from a 45 degree angle?
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.
What angle the gun can fire the maximum range?
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.
Is 200 degree an obtuse angle?
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.
A cannon shoots a projectile at an angle of 60 degrees above the horizontal with an initial speed of 30.0 ms. Calculate the maximum height of the projectile and its range?
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
Why there are two possiles angles for the same range of projectile?
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.
