h=u^2 sin^2x / 2g . where x is angle of release and h is the height of the projectile.
The max height depends only on the angle and speed at release. It doesn't depend on the projectile's weight.
45 degrees.
projection speed projection angle projection height
15.42 degrees
The range of projectile is maximum when the angle of projection is 45 Degrees.
It depends on the triangle. There is no description of this relationship that fits all triangles.
when a body is thrown at an angle in a projectile motion, the vertical component of the velocity is vcos(B) ..where v is the velocity at which the body is thrown and B represents the angle at which it is thrown.Similarly horizontal component is vsin(B). these components are useful in determining the range of the projectile ,the maximum height reached,time of ascent,time of descent etc.,
Range of a projectileThe path of this projectile launched from a height y0 has a range d.In physics, a projectile launched with specific initial conditions in a uniform gravity field will have a predictable range. As in Trajectory of a projectile, we will use:g: the gravitational acceleration-usually taken to be 9.80 m/s2 (32 f/s2) near the Earth's surfaceθ: the angle at which the projectile is launchedv: the velocity at which the projectile is launchedy0: the initial height of the projectiled: the total horizontal distance travelled by the projectileWhen neglecting air resistance, the range of a projectile will beIf (y0) is taken to be zero, meaning the object is being launched on flat ground, the range of the projectile will then simplify toso to increase the range θ shoud vary from 0 to 45 and after 45 it starts decreasing.
At 45° angle.
Get the value of initial velocity. Get the angle of projection. Break initial velocity into components along x and y axis. Apply the equation of motion .
The vertical component of the projectile's motion is uniformly accelerated, no matter what the angle of launch was.
The half maximum range of a projectile is launched at an angle of 15 degree