Calculating Time and Distance of a Baseball's Horizontal Motion at 161 km/h

[B-80]

A baseball leaves a pitcher's hand horizontally at a speed of 161 km/h. The distance to the batter is 18.3 m. Neglect air resistance.

(a) How long does it take for the ball to travel the first half of that distance?

161*.27777=44.7m/s

t=m/v=(18.3/2)/44.7=.2 seconds

wrong...

(b) How long does it take for the ball to travel the second half of that distance?
should be the same as (a) right?
(c) How far does the ball fall under gravity during the first half?
can't do without time
(d) How far does the ball fall under gravity during the second half?
can't do without time

 

[Astronuc]

0.2 s looks correct since the ball travels at constant velocity (neglecting air resistance).

Solve b with t = 0.2 s

and then solve c with the appropriate time 0.2 s later.

 

[kyle8921]

I would like to first clarify that the calculations provided in the original content are incorrect. The correct way to calculate the time for the ball to travel the first half of the distance is t=d/v, where d is the distance and v is the velocity. In this case, it would be t=18.3/44.7=0.41 seconds. The same applies for the second half of the distance, which would also take 0.41 seconds to travel.

To answer the remaining questions, we would need to take into account the acceleration due to gravity, which is 9.8 m/s^2. Since the ball is traveling horizontally, it is not affected by gravity in the horizontal direction. However, it will experience a vertical acceleration of 9.8 m/s^2 as it falls towards the ground.

To calculate the distance the ball falls under gravity during the first and second half of the distance, we can use the equation d=1/2at^2, where a is the acceleration and t is the time. For the first half, it would be d=1/2(9.8)(0.41)^2=0.84 meters. For the second half, it would also be 0.84 meters.

In conclusion, the time it takes for the ball to travel the first and second half of the distance is 0.41 seconds, and the distance it falls under gravity during each half is 0.84 meters. These calculations are based on the assumption that air resistance is neglected. If we were to take air resistance into account, the calculations would be more complex and require additional information.