The range of projectile is maximum when the angle of projection is 45 Degrees.
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.
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
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.
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.
45 degrees.
15.42 degrees
The half maximum range of a projectile is launched at an angle of 15 degree
At 45° angle.
45 degrees is the furthest one
Suppose a projectile is fired from a gun, we know that "g" remains constant and as we use horizontal component of velocity in range sov0 also remains constant. Only sin2θ responsible for change in range. The range will be maximum if sin2θ has its maximum value that is 1.for maximum range:sin2θ = 12θ = sin-1 (1)θ = 90/2θ = 45 (degree)therefor if projectile is projected with the angle of 45(degree) its range will be maximum.
Without considering any effects of air resistance and wind, the angle of projection that delivers the greatest horizontal range is 45 degrees above the horizon. Anything different ... either more or less than 45 degrees ... reduces the range.
Speed, weight, and shape of the projectile- along with the angle at which the barrel is raised above the horizonatal.
Factors include the angle at which it is fired, the speed of the projectile, its shape (ballistic coefficient) the air pressure and humidity.
A baseball, cannonball, or other projectile launched at a 45° angle above the horizon will achieve maximum horizontal range. A projectile launched straight up will achieve maximum altitude, but you kind of have to watch it and be careful when it returns.
"the higher the altitude the lower the range "
You cannot. You need to know either the initial speed or angle of projection (A).