Baseball player a bunts the ball by hitting it in such a way that it acquires an initial velocity of 2.4 m/s parallel to the ground. upon contact with the bat the ball is 1.2 m above the ground. player b wishes to duplicate this bunt, in so far as he also wants to give the ball a velocity parallel to the ground and have his ball travel the same horizontal distance as player a's ball does. however, player b hits the ball when it is 1.6 m above the ground. what is the magnitude of the initial velocity that player b's ball must be given

Let denote the position vector of the ball hit by player A. Then this vector has components

where is the magnitude of the acceleration due to gravity. Use the vertical component to find the time at which ball A reaches the ground:

The horizontal position of the ball after 0.49 seconds is

So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector of the ball hit by player B has

Again, we solve for the time it takes the ball to reach the ground:

After this time, we expect a horizontal displacement of 12 meters, so that satisfies

divide the right side by the left side

0.024÷1.2=0.02 for example

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