Playing shortstop, you pick up a ground ball and throw it to second base. The ball is thrown horizontally, with a speed of 25m/s , directly toward point A . When the ball reaches the second baseman 0.41s later, it is caught at point B.

What is the distance of vertical drop, AB?

Question

Playing shortstop, you pick up a ground ball and throw it to second base. The ball is thrown horizontally, with a speed of 25m/s , directly toward point A . When the ball reaches the second baseman 0.41s later, it is caught at point B.

What is the distance of vertical drop, AB?

anonymous · Answer

lets start with drawing to make obvious|dw:1401920410899:dw|
|dw:1401920810304:dw|
Calculate V y component by this formula
$$x = 0.5 a t ^{2} = 0.5* 9.8 * 0.41^{2}= 0.82369 m.$$
Calculate V x component ( it has nothing to affect it as we neglected air resistance).
$$y = Vt = 25* 041= 10.25 m.$$
and now by Pythagoras theorem.
$$ab = \sqrt{x ^{2}+y ^{2}}=\sqrt{10.25^{2}+0.82369^{2}}= 10.2830426 \approx 10.3 m$$
There you go.I hope you didn't get confused from distance x and y as I drew the reversely , so I used them as I drew.