A baseball leaves a pitcher's hand horizontally at a speed of 115 km/h. The distance to the batter is 18.3 m. (Ignore the effect of air resistance.) (a) How long does the ball take to travel the first half of that distance? (b) The second half? (c) How far does the ball fall freely during the first half? (d) During the second half?
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To solve this problem, we need to use the equations of motion for horizontal and vertical motion. In this case, we can assume the initial velocity in the vertical direction is zero since the ball is thrown horizontally.
(a) To find the time it takes for the ball to travel the first half of the distance, we can use the equation:
distance = velocity * time
Given that the distance is half of the total distance to the batter (18.3 m / 2 = 9.15 m) and the velocity is given in km/h, we need to convert it to m/s:
velocity = 115 km/h = (115 * 1000 m) / (3600 s) ≈ 31.9 m/s
Now we can rearrange the equation and solve for time:
time = distance / velocity
time = 9.15 m / 31.9 m/s ≈ 0.286 seconds
Therefore, it takes approximately 0.286 seconds for the ball to travel the first half of the distance.
(b) Since the ball is thrown horizontally and there is no horizontal acceleration, the time it takes for the ball to travel the second half of the distance will be the same as for the first half, which is approximately 0.286 seconds.
(c) To find the distance the ball falls freely during the first half, we can use the equation:
distance = 0.5 * acceleration * time^2
Since the ball falls freely, the acceleration is equal to the acceleration due to gravity, which is approximately 9.8 m/s^2. Substituting the values into the equation:
distance = 0.5 * 9.8 m/s^2 * (0.286 s)^2 ≈ 0.41 meters
Therefore, the ball falls freely for approximately 0.41 meters during the first half.
(d) Similarly, the distance the ball falls freely during the second half will also be 0.41 meters.