Answer:
94.13 ft/s
Explanation:
<u>Given:</u>
= time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
= distance to be moved by the rock long the horizontal = 98 yards
= displacement to be moved by the rock during the time of flight along the vertical = 0 yard
<u>Assume:</u>
= magnitude of initial velocity of the rock
= angle of the initial velocity with the horizontal.
For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

On dividing equation (1) by (2), we have

Now, putting this value in equation (2), we have

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.
The moon is talking to thecat at night that’s why
We can use the equation E = k | Q | r 2 E = k | Q | r2 to find the magnitude of the electric field. The direction of the electric field is determined by the sign of the charge,
<h3>What is electric and magnetic field ?</h3>
With the use of electricity and other types of artificial and natural illumination, invisible energy fields known as electric and magnetic fields (EMFs) and radiation are created.
- While the magnetic field is discernible by the force it exerts on other magnetic particles and moving electric charges, the electric field is actually the force per unit charge experienced by a non-moving point charge at any given location inside the field.
Learn more about Electromagnetic field here:
brainly.com/question/14372859
#SPJ4
Answer:

Acceleration, in m/s, of such a rock fragment = 
Explanation:
According to Newton's Third Equation of motion

Where:
is the final velocity
is the initial velocity
a is the acceleration
s is the distance
In our case:

So Equation will become:

Acceleration, in m/s, of such a rock fragment = 
it would be..... C
sorry if I am wrong I tryed to think, At least I try!