Physics, 25.09.2019 11:30, cameronking019

An object is thrown with an initial velocity v0 forming an angle θ with an inclined plane, which a in turn it forms an α-angle α with a horizontal plane.
a) find the distance x from the launch point to the point where the object falls. x = f (v0, θ, α, g)
b) find the distance for v0 = 20m / s, θ = 53o , α= 36 ° , g = 9.8m / s2

An object is thrown with an initial velocity v0 forming an angle θ with an inclined plane, which a i

Answers: 1

Show answers

Answers

answered: erik1franks

Refer to the figure shown below, which is based on the given figure.

d = the horizontal distance that the projectile travels.
h = the vertical distance that the projectile travels.

Part A
From the geometry, obtain
d = X cos(α) (1a)
h = X sin(α) (1b)

The vertical and horizontal components of the launch velocity are respectively
v = v₀ sin(θ - α) (2a)
u = v₀ cos(θ - α) (2b)

If the time of flight is t, then
vt - 0.5gt² = -h
or
0.5gt² - vt - h = 0 (3a)
ut = d (3b)

Substitute (1a), (1b), (2a), (2b) (3b) into (3a) to obtain
$0.5(9.8)( \frac{d}{u})^{2} -v_{0} sin(\theta - \alpha ) \frac{d}{u} - h = 0$
$4.9[ \frac{X cos \alpha }{v_{0} cos(\theta - \alpha } ]^{2} - v_{0} sin(\theta - \alpha ) [ \frac{X cos \alpha }{v_{0} cos(\theta - \alpha } ] - X sin \alpha = 0$
Hence obtain
$aX^{2}-bX=0 \\ where \\ a=4.9[ \frac{cos \alpha }{v_{0} cos(\theta - \alpha )}]^{2} \\ b = cos \alpha \, tan(\theta - \alpha ) + sin \alpha$
The non-triial solution for X is
$X= \frac{b}{a}$

$X= \frac{sin \alpha + cos \alpha \, tan(\theta - \alpha )}{4.9 [ \frac{cos \alpha }{v_{0} \, cos(\theta - \alpha )} ]^{2}}$

Part B
v₀ = 20 m/s
θ = 53°
α = 36°

sinα + cosα tan(θ-α) = 0.8351
cosα/[v₀ cos(θ-α)] = 0.0423

X = 0.8351/(4.9*0.0423²) = 101.46 m

X = 101.5 m
An object is thrown with an initial velocity v0 forming an angle θ with an inclined plane, which a i

An object is thrown with an initial velocity v0 forming an angle θ with an inclined plane, which a i

answered: Guest

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

so, refractive index of medium = 300000/200000 = 1.5

1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

∞ refractive index is usually for opaque mediums. so third medium is opaque medium.

answered: Guest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .