Answer:
1.) U = 39.2 m/s
2.) t = 4s
Explanation: Given that the
height H = 78.4m
The projectile is fired vertically upwards under the acceleration due to gravity g = 9.8 m/s^2
Let's assume that the maximum height = 78.4m. And at maximum height, final velocity V = 0
Velocity of projections can be achieved by using the formula
V^2 = U^2 - 2gH
g will be negative as the object is moving against the gravity
0 = U^2 - 2 × 9.8 × 78.4
U^2 = 1536.64
U = sqrt( 1536.64 )
U = 39.2 m/s
The time it takes to reach its highest point can be calculated by using the formula;
V = U - gt
Where V = 0
Substitute U and t into the formula
0 = 39.2 - 9.8 × t
9.8t = 39.2
t = 39.2/9.8
t = 4 seconds.
Answer:
The angle it subtend on the retina is
Explanation:
From the question we are told that
The length of the warbler is 
The distance from the binoculars is 
The magnification of the binoculars is 
Without the 8 X binoculars the angle made with the angular size of the object is mathematically represented as



Now magnification can be represented mathematically as

Where
is the angle the image of the warbler subtend on your retina when the binoculars i.e the binoculars zoom.
So

=> 

Generally the conversion to degrees can be mathematically evaluated as

Answer:
The normal line divides the angle between the incident ray and the reflected ray into two equal angles. The angle between the incident ray and the normal is known as the angle of incidence. The angle between the reflected ray and the normal is known as the angle of reflection.
Answer:
0.853 m/s
Explanation:
Total energy stored in the spring = Total kinetic energy of the masses.
1/2ke² = 1/2m'v².................... Equation 1
Where k = spring constant of the spring, e = extension, m' = total mass, v = speed of the masses.
make v the subject of the equation,
v = e[√(k/m')].................... Equation 2
Given: e = 39 cm = 0.39 m, m' = 0.4+0.4 = 0.8 kg, k = 1.75 N/cm = 175 N/m.
Substitute into equation 2
v = 0.39[√(1.75/0.8)
v = 0.39[2.1875]
v = 0.853 m/s
Hence the speed of each mass = 0.853 m/s