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.
Live long and prosper.
EDIT : Note that the following answer contains a small error --- the formula presented for the maximum height H lacks the divisor 2 that it should properly have. Only with its incusion could one obtain the general result that
tan θ = 4H/R.
The horizontal distance will be doubled.
90
its 45 degree
the equation is 1/2 x base x height or (base x height)/2
area=(1/2)*base*height
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!