Interesting question!
Answer: 75.96... degrees from the horizontal.
Let the projectile be launched with speed v at an angle θ degrees above the horizontal. Then its vertical speed component is v sin θ, and its horizontal component is v cos θ.
The time of flight is t = 2 v sin θ / g, so that the range R is given by
R = v t cos θ = 2 v^2 sin θ cos θ / g.
The maximum height is given by H = (v sin θ)^2 / (2 g).
(You may think of this is as merely an application of the standard result that, dropping from rest in the second half of the flight, the downward speed u = v sin θ must be related to the downward acceleration a and distance d travelled by u^2 = 2ad. In this context, where a = g and d = H, that is equivalent to the conservation of kinetic plus potential energy, of course.)
If R = H, then v^2 sin^2 θ / (2 g) = 2 v^2 sin θ cos θ / g.
Therefore sin θ / cos θ = tan θ = 4.
Thus θ = arctan (4) = 75.96... degrees.
The range of projectile is maximum when the angle of projection is 45 Degrees.
the angkle of projection is an angle and the projection
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
45 degrees to the horizontal will give the maximum flight time for a projectile. If a projectile was fired at 90 degrees to the horizontal, (straight upwards) the projectile will go straight upwards (ignoring the shape, form and aerodynamic properties of the projectile). Likewise if you were to fire a projectile at 0 degrees to the horizontal, the projectile would follow said course, IF gravity was not in effect; a projectile needs some form of vertical velocity to overcome gravity. Hence why 45 degrees will give you the longest distance and consequently flight time.
A cone shown in a circle
The range of projectile is maximum when the angle of projection is 45 Degrees.
45 degrees.
15.42 degrees
projection speed projection angle projection height
false....just by velocity the projection cannot be maximum.....for maximum projection the angle at which the projection is made and location would play a big role....ie..if two rockets are fired one from equator and one from pole with same velocity and same angle....the rocket fired from pole will have maximum projectile as it has to pass through less atmosphere hence less resistant....
"the higher the altitude the lower the range "
Max height H = u2 sin2@ / 2g So as we increase the angle of projection, then max height too increases and its value will be just u2/2g when it is projected vertically upwards ie @ = 90 deg
The half maximum range of a projectile is launched at an angle of 15 degree
At 45° angle.
h=u^2 sin^2x / 2g . where x is angle of release and h is the height of the projectile.
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 .
If you keep th velocity of projection and change the angle of projection from 75 degrees to 45 degrees what will happen to the horizontal distance the projectile travels? if you finish the nova net lesson you might learn the answer! It will travel a greater distance!