First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed

, and an accelerated motion on the y-axis, with initial speed

and acceleration

:


where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).
To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring

Therefore:

which has two solutions:

is the time of the beginning of the motion,

is the time at which the projectile hits the ground.
Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
<em>The answer you are looking for is: </em>
<em><u>True</u>
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Explanation:
Given that,
Electric field = 5750 N/C
Charge 
Distance = 5.50 cm
(a). When the charge is moved in the positive x- direction
We need to calculate the change in electric potential energy
Using formula of electric potential energy



Put the value into the formula


The change in electric potential energy is 
(b). When the charge is moved in the negative x- direction
We need to calculate the change in electric potential energy
Using formula of electric potential energy



Put the value into the formula


The change in electric potential energy is 
Hence, This is the required solution.
Answer:
6
Explanation:
Number of lines emanate from + 5 micro coulomb is 15 .
They terminates at negative charges that means at - 3 micro coulomb and - 2 micro Coulomb.
the electric field lines terminates at - 3 micro Coulomb and - 2 micro Coulomb is in the ratio of 3 : 2.
So the lines terminating at - 3 micro coulomb
= 
So the lines terminating at - 2 micro coulomb
= 
So, the number of filed lines terminates at - 2 micro Coulomb are 6.