Question

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, theball is 1.2 meters 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 hisball travel the same horizontal distance as player A’s ball does. However, player B hits the ball when it is 1.5 meters above the ground. What is the magnitude of theinitial velocity that player B’s ball must be given?

Answer 1

R1 = range of ball 1, R2 = range of ball 2

v1 = 1.9 m/s (velocity of ball 1), v2 = velocity of ball 2

h1 = 1.2 m (initial height of ball 1), h2 = 1.5 m (initial height of ball 2)

t1 = time it takes for ball 1 to fall, t2 = time it takes for ball 2 to fall

To start with, lets find out the time it will take for balls 1 and 2 to hit the ground...

Using constant acceleration kinematics, we find that:

and

solving for t1 and t2, we get:

and

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Now we can calculate the range of each ball...

Again, using basic kinematics (with no acceleration), we find the ranges to be:

R1 = v1 * t1 and R2 = v2 * t2

The problem specifies that R1 = R2, so we can set them equal to each other and solve for v2:

From this we can calculate that v2 = 1.7 m/s