To solve this problem we will apply the concepts related to energy conservation. With this we will find the speed before the impact. Through the kinematic equations of linear motion we will find the velocity after the impact.
Since the momentum is given as the product between mass and velocity difference, we will proceed with the velocities found to calculate it.
Part A) Conservation of the energy





Part B) Kinematic equation of linear motion,

Here
v= 0 Because at 1.5m reaches highest point, so v=0


Therefore the velocity after the collision with the floor is 3.7m/s
PART C) Total change of impulse is given as,





Answer:
Matter can not be created or destroyed, it only changes forms.
Explanation:
conservation of matter
(a)
is the wavelength in air of such a sound wave.
(b)
is the wavelength of this wave in tissue.
<u>Explanation:</u>
Frequency and wavelength can be related by the equation,
Velocity = Wavelength x Frequency

where,
v - velocity of light for all EM (electromagnetic) waves in vacuum
Given:
f - 4.50 MHz = 
a) To find the wavelength in air
We know,
Speed of sound in air = 343 m/s
Apply given frequency and speed of sound in air, we get

b) If the speed of sound in tissue is 1500 m/s, find the wavelength of this wave in tissue
Speed of sound in tissue, v = 1500 m/s

Answer:
Vf = 28 m/s
Explanation:
In order to find the final velocity of the rock, we will use the 3rd equation of motion. The third equation of motion for vertical direction is written as follows:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity = 9.8 m/s²
h = height dropped = 40 m
Vf = final velocity of the rock = ?
Vi = Initial Velocity of the rock = 0 m/s (since, rock was initially at rest)
Therefore,
(2)(9.8 m/s²)(40 m) = Vf² - (0 m/s)²
Vf = √(784 m²/s²)
<u>Vf = 28 m/s</u>