a tennis ball was held 4.5 ft above the surface of a tennis court and was thrown vertically downward with an initial velocity whose magnitude was 12 fps. Determine the magnitude of the velocity of the ball just before it struck the surface of the court

V^2 = Vo^2 + 2g*d.

Vo = 12 Ft/s.
g = 32FT/s^2.
d = 4.5 Ft.

Plug the given values into the Eq and
solve for V in Ft/s.

To determine the magnitude of the velocity of the ball just before it struck the surface of the court, we can use the principles of motion and gravity.

First, let's break down the given information:
- The height from which the tennis ball was dropped: 4.5 ft
- The initial velocity of the ball: 12 fps (feet per second)

The key equation we need to use is the equation of motion for an object in free fall:

v^2 = u^2 + 2as

Where:
- v is the final velocity of the ball
- u is the initial velocity of the ball
- a is the acceleration of the ball due to gravity
- s is the distance traveled by the ball

In this case, the ball is dropped vertically downwards, so the acceleration due to gravity will be -32.2 ft/s^2 (considering downward direction as positive).

Let's calculate the distance traveled by the ball (s):
Given that the ball was dropped from a height of 4.5 ft, the distance traveled will be equal to the initial height.

s = 4.5 ft

Now, let's calculate the final velocity (v) of the ball using the equation of motion:

v^2 = u^2 + 2as

v^2 = (12 fps)^2 + 2(-32.2 ft/s^2)(4.5 ft)

v^2 = 144 fps^2 - 289.8 fps^2

v^2 = -145.8 fps^2

Since the velocity cannot be negative, we take the square root of both sides:

v = √(-145.8) fps

As we cannot take the square root of a negative number in the real number system, the magnitude of velocity for the ball just before it strikes the surface of the court is 0 fps.

Therefore, the magnitude of the velocity of the ball just before it struck the surface of the court is 0 fps.