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.
45 degrees.
15.42 degrees
The half maximum range of a projectile is launched at an angle of 15 degree
At 45° angle.
The maximum range of a projectile is the distance it travels horizontally before hitting the ground. It is influenced by factors such as initial velocity, launch angle, and air resistance. In a vacuum, the maximum range is achieved at a launch angle of 45 degrees.
The maximum height of a projectile depends on its initial velocity and launch angle. In ideal conditions, the maximum height occurs when the launch angle is 45 degrees, reaching a height equal to half the maximum range of the projectile.
Changing the angle of projection affects the magnitude of range, maximum height, and time of flight. A higher angle will decrease the range and increase the maximum height while maintaining the time of flight. A lower angle will increase the range and decrease the maximum height while also maintaining the time of flight.
A projectile has maximum horizontal range when it is launched at an angle of 45 degrees to the horizontal. This angle allows for the ideal balance between the horizontal and vertical components of the projectile's velocity, ensuring that it travels the farthest distance before hitting the ground.
The launch angle and initial speed of a projectile are both factors that determine the range and height of the projectile. A higher launch angle with the same initial speed will typically result in a longer range but lower maximum height. Conversely, a lower launch angle with the same initial speed will result in a shorter range but a higher maximum height.
The range of a projectile is influenced by both the initial velocity and launch angle, while the height of the projectile is affected by the launch angle and initial height. Increasing the launch angle typically decreases the range but increases the maximum height of the projectile.
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.