Abaseball player standing on a platform throws a baseball out over a level playing field. the ball is released from a point 3.50 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. what horizontal distance will the ball travel before hitting the ground?

22.1 m

Explanation:

= initial speed of ball = 14.3 m/s

= Angle of launch = 27°

Consider the motion of the ball  along the vertical direction.

= initial speed of ball =

= acceleration due to gravity = - 9.8 ms⁻²

 = time of travel

 = vertical displacement = - 3.50 m

Using the kinematics equation that suits the above list of data, we have

Consider the motion of the ball along the horizontal direction.

= initial speed of ball =

 = Horizontal distance traveled

 = time taken = 1.74 s

Since there is no acceleration along the horizontal direction, we have

Let's answer this in a straightforward way:
The Electric field due to a point charge at a point with distance r relative to the point charge is:

For E1 we have is:

For E2=E1 we have:


We the get the following equation:


Because  we get from the above expression for q2: