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Grade: 11

                        

A baseball leaves the pitcher's hand horizontally at a speed of 92.0 mi/h, The distance to the batter is 60.0 ft. (a)How long does it take for the ball to travel the first 30.0 ft horizontally? The second 30.0 ft? (b) How far does the ball fall under gravity during the first 30 ft of its horizontal travel? (c) During the second 30.0 ft? (d) Why are these quantities not equal? Ignore the effects of air resistance.

5 years ago

Answers : (1)

Aditi Chauhan
askIITians Faculty
396 Points
							236-1598_1.PNG
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Therefore, the vertical distance travelled by the ball by the time it travels first 30.0 ft is 0.24 m .


236-1894_1.PNG
The vertical distance say travelled by in moving the last 30.0 ft is:
236-2236_1.PNG
Therefore, vertical distance travelled by in moving the last 30.0 ft is 0.70 m .
(d)
The vertical distance travelled by the ball is non-linear due to the presence of acceleration in the vertical direction whereas the distance travelled horizontally is independent of any acceleration, and remains constant over a fixed time interval. This is why the two distances are not equal, as one would expect it to be.
5 years ago
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