Answer:
x=2.4t+4.9t^2
Explanation:
This equation is one of the kinematic equations to solve for distance. The original equation is as follows:
X=Xo+Vt+1/2at^2
We know that the ball starts at rest meaning that its initial velocity and position is zero.
X=0+Vt+1/2at^2
Since it is going down the ramp, you can use the acceleration of gravity constant. (9.81 m/s^2) and simplify that with the 1/2.
X=Vt+4.9t^2
Note: Since the positive direction in this problem is down, you are adding the 4.9t^2, but if a question says that the downward direction is negative, you would subtract those values.
Now, substitute in your velocity value.
X=2.4t+4.9t^2
Answer:
a) 46.5º b) 64.4º
Explanation:
To solve this problem we will use the laws of geometric optics
a) For this part we will use the law of reflection that states that the reflected and incident angle are equal
θ = 43.5º
This angle measured from the surface is
θ_r = 90 -43.5
θ_s = 46.5º
b) In this part the law of refraction must be used
n₁ sin θ₁ = n₂. Sin θ₂
sin θ₂ = n₁ / n₂ sin θ₁
The index of air refraction is n₁ = 1
The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface
θ_s = 90 - θ
θ_s = 90- 43.5
θ_s = 46.5º
sin θ₂ = 1 / 1.68 sin 46.5
sin θ₂ = 0.4318
θ₂ = 25.6º
The angle with respect to the surface is
θ₂_s = 90 - 25.6
θ₂_s = 64.4º
measured in the fourth quadrant
Answer:
okay with you if you want to
