Launching angle of a projectile
Help with physics problem and solution
The initial speed vo of a projectile is 4 times the speed at its maximum height. What is the elevation angle at launching.
Solution:
The projectile launched at angle α and speed vo, has the vertical component of speed
vy = vosinα (1)
and the horizontal component
vx = vocosα (2)
_fig graph-launching-angle
At maximum height the vertical component vy = 0 and the only component of speed is the horizontal one, which is constant throughout the whole time of “flying”.
According to the content of the problem
vo/vocosα = 4
so
cosα = (1/4)
α = cos-1(0.25)
α = 75.5 deg
Notation cos-1 means the inverse trigonometric function and is defined as follows
cos-1(z)=β
if and only if cosβ = z
where
-1 ≤ z ≤ 1 and -π/2 ≤ β ≤ π/2
In other words
the inverse cosine of a number z is the angle whose cosine is z.
In some countries inverse trigonometric functions are written as
arccos(z), arcsin(z), arctan(z),or arctg(z)
The notation does not influence the definition.