Answer:
4.5 s, 324 ft
Explanation:
The object is projected upward with an initial velocity of

The equation that describes its height at time t is
(1)
where t, the time, is measured in seconds.
In order to find the time it takes for the object to reach the maximum height, we must find an expression for its velocity at time t, which can be found by calculating the derivative of the position, s(t):
(2)
At the maximum heigth, the vertical velocity is zero:
v(t) = 0
Substituting into the equation above, we find the corresponding time at which the object reaches the maximum height:

And by substituting this value into eq.(1), we also find the maximum height:

Answer:
The angle between the blue beam and the red beam in the acrylic block is

Explanation:
From the question we are told that
The refractive index of the transparent acrylic plastic for blue light is 
The wavelength of the blue light is 
The refractive index of the transparent acrylic plastic for red light is 
The wavelength of the red light is 
The incidence angle is 
Generally from Snell's law the angle of refraction of the blue light in the acrylic block is mathematically represented as
![r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ]](https://tex.z-dn.net/?f=r_F%20%3D%20%20sin%20%5E%7B-1%7D%5B%5Cfrac%7Bsin%28i%29%20%2A%20%20n_a%20%7D%7Bn_F%7D%20%5D)
Where
is the refractive index of air which have a value of
So
![r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ]](https://tex.z-dn.net/?f=r_F%20%3D%20%20sin%20%5E%7B-1%7D%5B%5Cfrac%7Bsin%2845%29%20%2A%20%201%20%7D%7B%201.497%7D%20%5D)

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as
![r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ]](https://tex.z-dn.net/?f=r_C%20%3D%20%20sin%20%5E%7B-1%7D%5B%5Cfrac%7Bsin%28i%29%20%2A%20%20n_a%20%7D%7Bn_C%7D%20%5D)
Where
is the refractive index of air which have a value of
So
![r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ]](https://tex.z-dn.net/?f=r_C%20%3D%20%20sin%20%5E%7B-1%7D%5B%5Cfrac%7Bsin%2845%29%20%2A%20%201%20%7D%7B%201.488%7D%20%5D)

The angle between the blue beam and the red beam in the acrylic block

substituting values


Answer:
Explanation:
Given that,
Mass of ball m = 2kg
Ball traveling a radius of r1= 1m.
Speed of ball is Vb = 2m/s
Attached cord pulled down at a speed of Vr = 0.5m/s
Final speed V = 4m/s
Let find the transverse component of the final speed using
V² = Vr²+ Vθ²
4² = 0.5² + Vθ²
Vθ² = 4²—0.5²
Vθ² = 15.75
Vθ =√15.75
Vθ = 3.97 m/s.
Using the conservation of angular momentum,
(HA)1 = (HA)2
Mb • Vb • r1 = Mb • Vθ • r2
Mb cancels out
Vb • r1 = Vθ • r2
2 × 1 = 3.97 × r2
r2 = 2/3.97
r2 = 0.504m
The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m
The required time,
Using equation of motion
V = ∆r/t
Then,
t = ∆r/Vr
t = (r1—r2) / Vr
t = (1—0.504) / 0.5
t = 0.496/0.5
t = 0.992 second