SOLUTION: A baseball player hit the ball with an initial velocity of 162kph and 45 degrees angle. A man is standing 250m from the player. How far must the man run in order for him to catch t

Question 1132491: A baseball player hit the ball with an initial velocity of 162kph and 45 degrees angle. A man is standing 250m from the player. How far must the man run in order for him to catch the ball? How fast must the man run for him to catch the ball?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A baseball player hit the ball with an initial velocity of 162kph and 45 degrees angle. A man is standing 250m from the player. How far must the man run in order for him to catch the ball? How fast must the man run for him to catch the ball?
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The horizontal and vertical speeds are equal when the ball is hit, and are 162 * sqrt(2)/2 = 81sqrt(2) km/hr = 22.5 m/sec
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162 km/hr = 45 meters/sec
Using 10 m/sec/sec for gravity, and zero (0) for the height when the ball is hit:
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height h(t) = -5t^2 + t*45sqrt(2)/2 meters
Find the time of flight:
The time to max height is the vertex of the parabola, at t = -b/2a
t = (-45sqrt(2)/2)/(-10) = 2.25sqrt(2) seconds
--> total flight time = 4.5sqrt(2) seconds
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Horizontal distance = 4.5sqrt(2)*22.5sqrt(2) = 202.5 meters
The fielder must run 250 - 202.5 = 47.5 meters to meet the ball.
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His speed is 47.5/(4.5sqrt(2)) m/sec =~ 7.464 m/sec
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