The fastest major league pitcher throws a ball at 45m/s. If he throws the ball horizontally, how far does it drops vertically on the 18.4m trip to the home plate?

time in air: 18.4/45 seconds

distance fell:1/2 g t^2

So why did you used the equation Vx=Vox-gt for time, instead of the equation X= -1/2gt^2 + Voxt + Xo. I used the second one because of they gave me 18.4 which is the distance traveled in the x-axes? thanks!

for time in air i got approx. .40s

To determine how far the ball drops vertically during its 18.4m trip to the home plate, we can make use of the equations of motion.

When the ball is thrown horizontally, its initial vertical velocity is 0 m/s. The only force acting on the ball in the vertical direction is the force due to gravity, which causes the ball to accelerate downward at a constant rate of 9.8 m/s^2.

Next, we can calculate the time it takes for the ball to travel the horizontal distance of 18.4m. Since the ball is thrown horizontally, there is no horizontal acceleration. Therefore, we can use the equation:

distance = velocity × time

Rearranging the equation to solve for time, we have:

time = distance / velocity

Substituting the given values, we have:

time = 18.4m / 45m/s = 0.408s

Now, we can determine how far the ball drops vertically during this time. We'll use the equation:

distance = (1/2) × acceleration × time^2

Substituting the values we know:

distance = (1/2) × 9.8m/s^2 × (0.408s)^2 = 0.79m

Therefore, the ball drops approximately 0.79 meters vertically during its 18.4-meter trip to the home plate.