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a baseball is hit at a height 1m and then caught at the same height. It travels alongside a wall, moving up past the top of the wall 1.00 s after it is hit and then down past the top of the wall 4.00 s later, at distance 50m farther along the wall. Please explain. (a) What horizontal distance is traveled by the ball from hit to catch? (b) magnitude (c) angle (relative to the horizontal) of the ball's velocity just after being hit? (d) How high is the wall?

a baseball is hit at a height 1m and then caught at the same height. It travels alongside a wall, moving up past the top of the wall 1.00 s after it is hit and then down past the top of the wall 4.00 s later, at distance 50m farther along the wall.  Please explain.


(a) What horizontal distance is traveled by the ball from hit to catch?


(b) magnitude


(c) angle (relative to the horizontal) of the ball's velocity just after being hit?


(d) How high is the wall?

Grade:11

1 Answers

Mukul Shukla
40 Points
10 years ago

Let us assume that the ball was hit with velocity V making an angle 'b' with the horizontal.

So it horizontal component of velocity will be constant and will be equal to v cos b.

Now we know the ball took one sec to move pas the wall and then three more seconds to move the distance when it was above the wall. So it must have taken one more second till it was caught; making the total time interval to 5 seconds.

Also we know the ball travelled 50m in three seconds with a constant horizontal velocity; there fore the horizontal component must be equal to 50/3 m/sec.

Therefor total horizontal distance will be equal to 50/3*5= 250/3 m.

Now if horizontal component is 50/3=v cos b then v sin b= say x.. this will give you the height of the wall as this velocity took one sec to cover the height of the wall.

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