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Problem 28 Hard Difficulty

Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.9 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 1.5 m above the ground. What is the magnitude of the initial velocity that player B’s ball must be given?

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Video Transcript

so the question states that two batters are bunting a ball and both batters want their ball to go the same distance. The first batter hits his ball at 1.9 meters per second in the horizontal direction at starting at a height of 1.2 meters and the second player hits a ball at Speed V at a height of 1.5 meters. And we're trying to find what this Speed V has to be so that they could go the same distance. Um so the first thing we have to do is figure out how far the ball actually goes. So what this distance X is, And to do this, we first need to find how long the ball is in the air for the first part of our for the first player. So we can use our kingdom attics formula, which states that the change in displacement is equal to the initial lost e times time plus 1/2 a T squared. We know that in the vertical direction, the initial velocity, he's going to zero. So this term cancel and we know the displacement in the vertical direction is 1.2 meters for the first player and we know the acceleration will be 9.8 because of gravity. So we can solve for the time that the ball will be in the air and the so give us point for 95 seconds. Now that we know the time that the ball is in the air, we can figure out how far the ball goes because we know the horizontal lost the ball so we can use the equation. Delta X is equal to be not t. We know the horizontal velocity in the time. So when we plug this in, get Delta X is equal to 1.9 meters per second times point for 95 seconds, which means the distance travelled by the ball is 0.941 meters. So now that we know the distance that the ball travels, we can tailor this to our second player to figure out how far there to figure out what the velocity needs to be, um, for the ball to travel the same distance. So it's pretty much the same thing that we did in for Player one. We need to first find the time it takes for the ball to reach the ground so we can use our and the equation that we used first player. So we know the vertical displacement is 1.5 meters, and we know that this term will cancel out because there's no vertical lost e. Um and once again, our acceleration is 9.8 meters per second squared. We can solve for the time and that gives us t is equal 2.5 53 seconds. And now that we know the time, we can plug it into this formula once again that says that felt it exit B t. And this is for our horizontal velocity to we know that the ball is going to give 0.941 meters. We're trying to find the horizontal velocity and we know the time it takes for the ball to reach the ground, which is 5.53 seconds. And so when we saw for ve we get that V is equal to 1.7 years per second

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John D. Cutnell, Kenneth W. Johnson

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Video Transcript

John D. Cutnell, Kenneth W. Johnson

Physics

Christina Krawiec

Andy Chen

Farnaz Mohseni

Jared Enns

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Physics

Christina Krawiec

Andy Chen

Farnaz Mohseni

Jared Enns

John D. Cutnell, Kenneth W. JohnsonPhysics

Christina Krawiec

Andy Chen

Farnaz Mohseni

Jared Enns

John D. Cutnell, Kenneth W. Johnson

Physics