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 answer these questions, we need to calculate the time it takes for the baseball to travel half the distance, and the distance it falls freely during that time.
(a) To find the time it takes for the ball to travel the first half of the distance, we can use the formula:
time = distance / speed
Given the distance of 18.3 m and the speed of 115 km/h, we need to convert the speed to m/s:
115 km/h = 115000 m / (3600 s)
Plugging in the values:
time = 18.3 m / (115000 m / (3600 s))
Simplifying:
time = 18.3 m x (3600 s) / 115000 m
Calculate this value to find the time it takes for the first half.
(b) The second half of the distance will also take the same time as the first half, assuming the initial speed remains constant.
(c) To find how far the ball falls freely during the first half, we can use the formula:
distance = 0.5 x gravity x time^2
Since the ball is thrown horizontally, the vertical movement is due to gravity. We know that the initial vertical velocity is zero, so the equation simplifies to:
distance = 0.5 x 9.8 m/s^2 x time^2
Plug in the calculated time value from part (a) to find the distance fallen during the first half.
(d) The ball falls freely for the same amount of time during the second half as it does during the first half. Therefore, the distance fallen during the second half is the same as in part (c).
Solve these equations to find the answers to the questions.