To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as

Where,
= Mass of each object
= Initial velocity of each object
= Final Velocity
Since the receiver's body is static for the initial velocity we have that the equation would become



Therefore the velocity right after catching the ball is 0.0975m/s
Answer:
D I think I might be wrong its been a while scense I did something like that
Answer:
The speed after being pulled is 2.4123m/s
Explanation:
The work realize by the tension and the friction is equal to the change in the kinetic energy, so:
(1)
Where:

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.
Additionally, the kinetic energy is equal to
, so if the initial velocity
is equal to zero, the initial kinetic energy
is equal to zero.
Then, replacing the values on the equation and solving for
, we get:


So, the speed after being pulled 3.2m is 2.4123 m/s
Answer:
a)
Y0 = 0 m
Vy0 = 15 m/s
ay = -9.81 m/s^2
b) 7.71 m
c) 3.06 s
Explanation:
The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards
Y(0) = 0 m
Vy(0) = 15 m/s
ay = -9.81 m/s^2 (negative because it points down)
Since acceleration is constant we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
To find the highest point we do the first time derivative (this is the speed:
V(t) = Vy0 + a * t
We equate this to zero
0 = Vy0 + a * t
0 = 15 - 9.81 * t
15 = 9.81 * t
t = 0.654 s
At this time it will have a height of:
Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m
The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.
0 = Y0 + Vy0 * t + 1/2 * a * t^2
0 = 0 + 15 * t - 1/2 * 9.81 t^2
0 = 15 * t - 4.9 * t^2
0 = t * (15 - 4.9 * t)
t1 = 0 This is the moment it jumped into the air
0 = 15 - 4.9 * t2
15 = 4.9 * t2
t2 = 3.06 s This is the moment when it falls again.
3.06 - 0 = 3.06 s