Mass have no effect for the projectile motion and u want to know the height "h"
first,
find the vertical and horizontal components of velocity
vertical component of velocity = 12 sin 61
horizontal component of velocity = 12 cos 61
now for the vertical motion ;
S = ut + (1/2) at^2
where
s = h
u = initial vertical component of velocity
t = 0.473 s
a = gravitational deceleration (-g) = -9.8 m/s^2
h=[12×sin 610×0.473]+[−9.8×(0.473)2]
u can simplify this and u will get the answer
h=.5Gt2
H=1.09m
The slowest line is the solid line and the fastest is the dotted line that crosses the solid line
for future reference you just need to find the slope or the line which is traveling most vertical
It probably is the actual answer.
Answer:
D = 2.38 m
Explanation:
This exercise is a diffraction problem where we must be able to separate the license plate numbers, so we must use a criterion to know when two light sources are separated, let's use the Rayleigh criterion, according to this criterion two light sources are separated if The maximum diffraction of a point coincides with the first minimum of the second point, so we can use the diffraction equation for a slit
a sin θ = m λ
Where the first minimum occurs for m = 1, as in these experiments the angle is very small, we can approximate the sine to the angle
θ = λ / a
Also when we use a circular aperture instead of slits, we must use polar coordinates, which introduce a numerical constant
θ = 1.22 λ / D
Where D is the circular tightness
Let's apply this equation to our case
D = 1.22 λ / θ
To calculate the angles let's use trigonometry
tan θ = y / x
θ = tan⁻¹ y / x
θ = tan⁻¹ (4.30 10⁻² / 140 10³)
θ = tan⁻¹ (3.07 10⁻⁷)
θ = 3.07 10⁻⁷ rad
Let's calculate
D = 1.22 600 10⁻⁹ / 3.07 10⁻⁷
D = 2.38 m
